Chap2Kinematics fundamental

合集下载

The Kinematic Composition of MgII Absorbers

a rXiv:as tr o-ph/9712235v117Dec1997Accepted by the Astrophysical Journal The Kinematic Composition of Mg ii Absorbers 1Jane C.Charlton 2and Christopher W.Churchill Astronomy and Astrophysics Department Pennsylvania State University,University Park,PA 16802email:charlton,cwc@ Accepted for publication:Astrophysical Journal ABSTRACT The study of galaxy evolution using quasar absorption lines requires an understand-ing of what components of galaxies and their surroundings are contributing to the absorption in various transitions.This paper considers the kinematic composition of the class of 0.4<z <1.0Mg ii absorbers,particularly addressing the question of what fraction of this absorption is produced in halos and what fraction arises from galaxy disks.We design models with various fractional contributions from radial infall of halo material and from a rotating thick disk component.We generate synthetic spectra from lines of sight through model galaxies and compare the resulting ensembles of Mg ii pro-ﬁles with the 0.4≤z ≤1.0sample observed with HIRES/Keck.We apply a battery of statistical tests and ﬁnd that pure disk and pure halo models can be ruled out,but that various models with rotating disk and infall/halo contributions can produce an ensemble that is nearly consistent with the data.A discrepancy in all models that we considered requires the existence of a kinematic component intermediate between halo and thick disk.The variety of Mg ii proﬁles can be explained by the gas in disks and halos of galaxies not very much diﬀerent than galaxies in the local Universe.In any one case there is considerable ambiguity in diagnosing the kinematic com-position of an absorber from the low ionization high resolution spectra alone.Future data will allow galaxy morphologies,impact parameters,and orientations,Fe ii /Mg ii of clouds,and the distribution of high ionization gas to be incorporated into the kinematic bining all these data will permit a more accurate diagnosis of the physical conditions along the line of sight through the absorbing galaxy.Subject headings:quasars:absorption lines —galaxies:structure —galaxies:evolution1.IntroductionQuasar absorption lines(QALs)provide a diagnostic of the gaseous conditions in and around galaxies as they form and evolve.Most Mg ii absorbers at z<1are produced within40h−1kpc of known>0.1L∗galaxies(Steidel1995).High resolution spectroscopy(HIRES/Keck)shows that they have multiple absorbing components(Churchill,Vogt,&Charlton1998,hereafter CVC98). These components should provide information about how the Mg ii gas is distributed spatially and kinematically.More generally,multiple component structure is apparent in various ionization stages of the diﬀerent chemical elements probed by QALs over the history of the Universe.Before QALs can be fruitfully applied to a detailed study of conditions in galaxies we must resolve many of the remaining ambiguities inherent in their interpretation.Based upon low resolution spectra,the Mg ii absorbers with W(Mg ii)>0.3˚A have generally been interpreted as material infalling into the halos of the normal>0.1L∗galaxies(Steidel1995; Mo&Miralda–Escud´e1996).However,most L∗galaxies are disk galaxies,and the disks themselves contain the H i necessary for ionization conditions that allow Mg ii to survive.Disks in the local Universe clearly extend well beyond the optical radii(Irwin1995),but it is hard to establish directly the H i distribution at N(H i)<1019cm−2.In a few cases,sensitive21cm measurements provide maps of galaxy disks down to N(H i)=1018cm−2,and in these cases the disks extend to tens of kpc (Corbelli&Salpeter1993;van Gorkom et al.1993).In the M81group,interactions lead to large N(H i)in aﬂattened distribution well beyond50kpc of M81itself(Yun,Ho,&Lo1994),but some relatively isolated dwarfs have been found to be extended also(Hoﬀman et al.1993).As shown in Charlton and Churchill(1996,hereafter CC96),for a disk with larger scale–height outer regions, the random orientation cross–section is competitive with that of spherical halos.In fact,CC96 argue that after possible biases in the available sample of Mg ii absorbers are taken into account, models can be designed that are consistent with either a spherical or a thick disk geometry.Simply because of the cross–section of known H i disks,galaxy disks must make some contri-bution to Mg ii absorption.Also,some fraction of absorbing galaxies are diskless(ellipticals),so at minimum there must be contribution to the absorption by these two kinematic components.Here we address the question of what the dominant contribution is to Mg ii absorption.Is the larger fraction of the Mg ii column density from disk material or from halo material?Does the answer to this question vary from galaxy to galaxy or from one line of sight to another within a single galaxy?Based upon the absorption properties along lines of sight through the Milky Way and through nearby galaxies,we can begin to infer the answers to these questions at z=0.Lines of sight looking out from our special vantage point in the Milky Way pass mostly through disk material,and from the ratios of various transitions and their positions in velocity along the line of sight,the nature of the absorbing clouds can be inferred as H i regions,H ii regions,superbubbles,or high velocity halo clouds(Spitzer&Fitzpatrick1993;Fitzpatrick&Spitzer1994;Spitzer&Fitzpatrick1995;Welty et al.1997).As with these many Galactic sight–lines,at higher redshifts there is considerable kinematic variation in the proﬁles.Churchill,Steidel,and Vogt(1996,hereafter CSV96)showthat the variations in the HIRES/Keck proﬁles of0.4<z<1.0Mg ii absorbers are not strongly correlated with the impact parameter,luminosity,or morphological type of the identiﬁed absorbing galaxy.The large scatter in the relationships between absorption and galaxy properties could be produced by variations due to clumpiness and discrete structures that lead to variations within the individual galaxies.The variations due to clumpy structures will complicate eﬀorts to extract kinematic informa-tion from individual proﬁles.Earlier studies of kinematic signatures demonstrated the basic proﬁle shapes expected from various spatial and kinematic laws(Weisheit1978;Lanzetta&Bowen1992). Lanzetta and Bowen suggested that a trend might exist where rotating disk signatures are charac-teristic of small impact parameter Mg ii absorbers,while“double–horned”infall proﬁles arise in the outer halos of the galaxies.The scatter observed in the relationships between impact parameter and absorption proﬁle properties indicates that the situation is more complicated(CSV96).There is hope that a given population of absorbers will give rise to proﬁles that reﬂect the underlying kinematic and spatial distributions in their host galaxies,but it is important to consider the eﬀect of stochastic variations on these proﬁles.Prochaska and Wolfe(1997)showed synthetic proﬁles produced by clouds selected from a thick exponential rotating disk and found that this kinematic law is consistent with the properties of high redshift damped Lyαabsorbers.In this paper we compare the properties of an ensembles of proﬁles drawn from various kine-matic models(and combinations of kinematic models)to the ensemble of observed HIRES/Keck proﬁles of0.4<z<1.0Mg ii absorbers.The following four questions motivate our simulations.1.Is it possible to extract enough information from a particular proﬁle to identify it as associatedwith a particular kind of galaxy,a line of sight through a particular part of a galaxy,a galaxy undergoing some particular stage of formation or evolution,etc.?To what extent will we be able to resolve the kinematic ambiguities?2.Can the statistical properties of the observed ensemble be reproduced by a population ofdisks and/or halos of galaxies?The Mg ii absorbing galaxies at0.4<z<1.0cover all morphological types and the range of luminosities down to0.1L∗.Are there proﬁles that require unusual kinematic laws or are all of them consistent with what we would expect for lines of sight through the population of galaxies at z=0?3.Can the variety of observed absorption proﬁles be reproduced within the context of a singlekinematic model,or a weighted combination of two(such as disk and halo)?Can variations in the proﬁles be due to the expected line of sight diﬀerences through one class of galaxies or are the proﬁles too varied?4.In view of future studies,what role will kinematic analysis of high resolution absorptionproﬁles of low ionization gas play in broader studies that also include galaxy properties, Lyα,and higher ionization gas?By combining this information,will we be able to extractinformation about the conditions along a sight–line through a particular galaxy,and about the global properties of the galaxy at the same level as is done through the Galaxy?The answers to these questions may lead to a characterization of the population of Mg ii ab-sorbers at0.4<z<1.0in terms of their kinematic composition.In this redshift range the absorbing galaxy can more readily be identiﬁed and studied,providing more leverage on the interpretation of absorption properties.Lines of sight through N–body hydrodynamic simulation boxes(which also model the ionization states of various metals)show that diﬀerent kinematic signatures result from galaxies at diﬀerent stages of evolution(Rauch,Haehnelt,&Steinmetz1997).A population of absorbers could be dominated by the kinematics of material that has just separated from the Hubbleﬂow,material falling into a galaxy halo,or material in a well–formed disk.Establishing the contributions of spatial and kinematic components to Mg ii proﬁles at low to intermediate redshift is prerequisite to interpreting their evolution as the redshift range of observability increases through studies in the UV and near–IR(Churchill1997b).The second section of this paper presents the26observed0.4<z<1.0Mg ii proﬁles and discusses the selection and analysis of this sample.Even a qualitative examination of these proﬁles suggests certain interpretations of their kinematic composition.Guided by these interpretations and by properties of nearby galaxies we designed kinematic models.The details of the model construction are outlined in§3,which also describes our statistical comparisons of the observed and model spectra.The results of our analysis is given in§4,where we address which of the kinematic models are formally consistent or inconsistent with the ensemble of observed spectra. In§5,we return to the four questions posed in this introduction and assess the extent to which we can characterize the kinematic and spatial distribution of the population of0.4<z<1Mg ii absorbers.2.HIRES/Keck Observations of0.4<z<1.0Mg ii AbsorbersThe observed sample for comparison to kinematic models has been selected from the larger study of Mg ii absorbers with equivalent width W(2796)>0.3˚A(Churchill1997a;CVC98).Al-though the full data set included systems out to z=1.8,for two reasons we choose to limit this study to the z<1Mg ii absorbers.First,this sample is unbiased in equivalent width below this redshift cutoﬀ,while several of the higher redshift systems were selected to be particularly strong. More importantly,there could be an evolution in the kinematic properties of the global population of galaxies over the larger redshift interval0.4<z<1.8.The redshift interval of0.4<z<1.0 covers the lookback time∼4−9Gyr.The data were obtained with the HIRES spectrometer(Vogt1994)on the Keck I telescope and have a spectral resolution of45,000,corresponding to6.6km s−1(Churchill1997a;CVC98).In addition to the Mg iiλλ2976,2803doublet,Mg i(2853)and several Fe ii transitions were observed. These prove to be important for the study since they allow more accurate Voigt proﬁleﬁtting,witha more realistic number of subcomponents in cases where the Mg ii doublet is saturated.So that we have a more uniform sample for the kinematic study we also eliminate all systems for which the5σequivalent width detection limit for Mg ii(2796)is greater than0.02˚A in the rest frame.In Figure1,we display only the Mg ii(2796)transition for each of the26systems,with Voigt proﬁle ﬁts determined using all available Fe ii,Mg i,and Mg ii transitions.What follows in this paper is a quantitative comparison of the data to synthetic spectra from the kinematic models.However,visual inspection of the HIRES/Keck Mg ii proﬁles in Figure1 already leads us to two simple conclusions about the kinematics of these objects.First,many systems are dominated by a strong blended component with a width of40–100km s−1.This range is not consistent with the velocity spread that we would expect for halo kinematics,whether it be infall,outﬂow,or random isotropic motion(for an example of proﬁles from infall halo kinematics see Figure5).Second,in some proﬁles there are outlying weaker components spread over a larger∆v of hundreds of km s−1.These large spreads are clearly inconsistent with the kinematics expected for a rotating disk,even when random vertical motions are included(see Figure4for examples of disk kinematics).Thus,simply based upon qualitative arguments,it is immediately apparent that some combination of disk and halo kinematics is needed to produce the observed ensemble of proﬁles.This could require either some galaxies that are pure disk galaxies and others that are pure halo galaxies,or it could imply that some proﬁles require both a disk and a halo component in the responsible galaxy.3.The Models and Their Simulated SpectraWe consider models with rotating disk components and with radial infall/halo components. The model disks are extended–we assume a radius of1.5R(L∗K)where R(L∗K)=35(L/L∗K)0.15kpc in order that the population of inclined disks can be consistent with the frequency with which galaxies within R(L∗K)of quasar lines of sight produce absorption(CC96).Halos are assigned a radius of R(L∗K).There areﬁve“populations”of kinematic/geometric models that we present in this study:1)a single population of pure disk models;2)a single population of pure halo models;3)a“two–population”model,in which50%of galaxies are pure disks and50%are pure halos;4)a single population of“hybrid”models,in which every galaxy has75%of its gaseous clouds in a disk and25%of its clouds in a halo(75D/25H);and5)a single population of“hybrid”models,in which every galaxy has50%of its clouds in a disk and50%of its clouds in a halo(50D/50H).In the course of our study,we investigated two–population models with various fractions of disk/halo galaxies,but we present only the50D/50H case for purposes of illustration.We also investigated other hybrid models,each with a diﬀerent fraction of disk/halo clouds.Again,we show only two of these cases in this paper,but draw upon the general study for our conclusions.The full problem of the physical conditions of the clouds is presently intractable.Our goal is simply to test the kinematic distribution of absorbing clouds and to rule out certain classes of mod-els,if possible(we make no attempt to diagnose the cloud chemical abundances or photoionization properties).Thus the individual cloud properties are selected from the measured distributions of column densities and Doppler parameters(Churchill1997a;CVC98).More precisely,we draw from underlying distributions that produce proﬁles consistent with the observed distributions once the simulated spectra are analyzed in the exact same way as were the data.The basic simulation procedure is as follows.For each simulated line of sight we pick a galaxy from the Schechter luminosity function withα=−1.Its rotation or radial infall velocity is given by the Tully–Fisher relation.Spherical clouds are placed in a spatial distribution and each of these clouds is assigned a velocity based upon a simple kinematic.Each cloud is assigned a Mg ii column density selected from a power law distribution and a Doppler parameter selected from a Gaussian with a low–end cutoﬀ.The“best”power law exponent and Doppler parameter spread and cutoﬀdepends upon the amount of blending in a particular model.3Once the model galaxy is deﬁned, we run a line of sight through it at impact parameter chosen by area weighting considerations and determine which clouds are ing the velocities,column densities,and Doppler parameters of these clouds,we generate simulated spectra.These spectra are then convolved with the HIRES instrumental proﬁle,sampled with the pixelization of the HIRES CCD,and degraded to have signal to noise S/N ratios consistent with the data.The S/N of any given simulated spectrum is chosen from the S/N distribution of the data.We generate the Fe ii proﬁles also, assuming thermal scaling of the Doppler parameters,and applying the observed relationlog N(Fe ii)≃log N(Mg ii)−0.3,(1) for the column densities.Thermal scaling is consistent with the observed ratios of the Mg ii and Fe ii Doppler parameters(Churchill1997a;CVC98).Having followed this procedure we have a set of spectra very similar to the observations and we can proceed to analyze them in precisely the same manner.The proﬁleﬁtting procedure used in our analysis is a two step process.Theﬁrst step is an application of the automated Voigt proﬁleﬁtter,autofit(Dav´e et al.1997),which we have modiﬁed to account for broadening due to the instrumental spread function.autofit was applied to the Mg ii(2796)transition and this solution was scaled to give parameters for the Voigt proﬁles to model the Mg ii(2803)and Fe ii transitions.This initial model was reﬁned using minfit(Churchill 1997a),a maximum likelihood least–squaresﬁtter.MINFIT minimizes theχ2between the data and the Voigt proﬁle model by adjusting the column densities,Doppler parameters,and number of Voigt proﬁle components(with the goal of minimizing the number of these components).3.1.Detailed ProcedureFor infall/halo models we place the clouds randomly within the spherical distribution of radius R(L K).The magnitude of each cloud velocity is set equal to the value given by the Tully–Fisher relation and its direction is radial towards the center of the sphere.In practice,each cloud is taken to have a radius of1kpc,but this is rather arbitrary since a cloud radius which is smaller would then require a larger number of clouds per halo such that the number of intercepted clouds remains conserved at a number consistent with the data.Thus we describe each model by its integrated cross sectional area,Nσcl,equal to the number of clouds times the cross section of each.For disk models there are a larger number of parameters and assumptions.The disk is taken to have a full height of1kpc out to a radius of10kpc,increasing linearly to a10kpc height at its maximum radius of1.5R(L K).The clouds are rotating diﬀerentially with aﬂat rotation curve with velocity given by the Tully–Fisher law.In addition each cloud is assigned a velocity in the direction perpendicular to the disk,selected from a Gaussian distribution withσ=15or25km s−1.For a given kinematic model we mustﬁnd the location in the parameter space of cloud property distributions that results in spectra that are consistent with the observed column density and Doppler parameter distributions.The parameters that describe the input cloud distributions are the slope of the Mg ii column density distribution,and the Gaussianσand lower cutoﬀof the Doppler parameter distribution.The column density distribution has a lower cutoﬀof log N(Mg ii)= 11.3cm−2for all grid points.This is the minimum cloud that could be detected for the highest S/N spectrum in our sample.It is not practical toﬁt all proﬁles for each grid point,thus we have devised a test that relies on theﬂux distribution of all pixels within absorption features.In Figure2,this test is illustrated for a small region of one of our model grids,which includes the cloud properties and the number of clouds in the galaxy as parameters.In fact,because of the large number of pixels in the distributions,this test provides a sensitive method for reﬁning the grid.In order toﬁnd a compatible model,it was critically important to give the model spectra the same distribution of S/N as the data.From a typical grid of24–48choices of parameters(for a given kinematic model)we considered only models that showed a probability of more than1% by the Kolmogornov–Smirnov test to be drawn from the same pixelﬂux distributions as the data. Typically,this techniques allowed us to exclude all but3–5grid points,which we then proﬁleﬁt and conducted further statistical comparisons.In Table1,we list the parameter choices that yielded adequateﬁts for various classes of models,along with their KS probabilities.Typically,the eﬀect of blending can be compensated by a steeper power law for the N(Mg ii)distributions,since the lowestﬂux pixels are often the result of the combined eﬀects of several lines.Increasing the vertical velocity dispersion of the disk kinematics decreases the number of saturated pixels.An increase in Nσcl can be compensated by a decrease in the mean Doppler parameter,but ultimately the output Doppler parameter distribution fromﬁtting must match the observed Doppler parameter distribution.From a maximum likelihoodﬁt to the observed distribution of0.4≤z≤1.0Mg ii absorbercolumn densities,we obtain power–law slope ofδ=1.74to n(N)∼N−δwith lower cutoﬀlog N= 12.5cm−2.In fact,for models withﬂux distributions consistent with the data,we found that the slope of the output column density distribution was similar to the input slope,over a reasonable ﬁtting range.Therefore an input slope ofδ=1.74was suitable.The Doppler parameter distribution of observed data,excluding values with fractional errors greater than unity,is bestﬁt by a Gaussian truncated at2–3km s−1,with a peak of3.5km s−1and a standard deviation of3–4km s−1. An input Doppler parameter distribution with a mean of3–4km s−1,aσof2km s−1,and a lower cutoﬀof2km s−1led to a goodﬁt of the output distribution with the observed data.The output distributions of column densities and Doppler parameters for a consistent hybrid model with50D/50H contributions are compared to the input and observed data distributions in Figure3. Based on these considerations we select the following models listed in Table1for further analysis: Pure Disk2,75D/25H Hybrid2,50D/50H Hybrid2,Pure Infall1,and a two–population mix in which half the systems are Pure Infall1and half are Pure Disk23.2.Statistical AnalysisStatistical tests fall into three categories:(1)Pixel by pixel comparisons:These tests involve the distribution of the number of pixels or pixel pairs satisfying various criteria.These have the advantage of large number statistics so that F–tests and KS tests are very sensitive to diﬀerences in model distributions.This is why the pixelﬂux distribution was judged an eﬀective method to reﬁne the model grid in the previous section.The disadvantage of these types of tests is that they average together the diﬀerences between individual systems,and thus they do not test whether a model can explain the variation in properties from system to system.(2)System by system statistics:The test compare the proﬁle shape and the Voigt proﬁle model cloud properties system by system.Examples of these statistics are the various moments of the proﬁles of entire systems(weighted by the apparent optical depth)and the number of clouds per system.These test are subject to small number statistics;comparisons can only be so good as the statistics on the26proﬁles in the observed sample permit.(3)Cloud or subfeature statistics:These tests are in some ways intermediate between the previous two.A subfeature is deﬁned as a detected region of a spectrum that is separated from other detected pixels by at least2.5HIRES resolution elements.An example of a subfeature statistic is the number of subfeatures in a given velocity bin.Cloud properties are parameterized by the Voigt proﬁleﬁt column densities and Doppler parameters.An example of a cloud statistic is the cloud–cloud two point clustering function.These statistics combine the individual systems and thus average out variations between them like(1).Although they are subject to small number statistics,there are more subfeatures and clouds than there are systems,so they are somewhat better in that sense than(2).Many plethora of these tests were performed on all models and on the data.For presentation we choose a range of tests that best distinguish the various diﬀerences that could exist between our models.The following battery was selected:(A)∆V of pixel pairs for which both pixels fall in the sameﬂux bin0.0–0.2,0.2–0.4,0.4–0.6,0.6–0.8,and0.8–1.0;(B)Histogram of number of clouds per system;(C)Histogram of optical depth weighted velocity widths,ωv,for systems,whereω2v= v2v1τa(v)(v− v )2dv/ v2v1τa(v)dv,(2) andτa is the apparent optical depth deﬁned byτa(λ)=ln I c(λ)4.Results of Kinematic ModelsSample proﬁles of the models of various types,selected from the grid in Table1,are presented in Figures4–8.These models are most consistent with the pixelﬂux distribution as well as with the observed column density and Doppler parameter distributions.Illustrated are27proﬁles each for:the Pure Disk models(Figure4);the Pure Infall models(Figure5);the75D/25H Hybrid models(Figure6);the50D/50H Hybrid models(Figure7);and,the50/50two–population models (Figure8).The50/50two–population models shown in Figure8are drawn from the models presented in Figures4and5.This population illustrates the ambiguity that could exist in classifying any given observed system.4.1.Pixel by Pixel StatisticsWe adapted a procedure developed in Cen et al.(1997)and considered the pairwise velocity diﬀerences for pixels in variousﬂux bins and within a system.We also examined the distributions of velocity relative to the velocity zero point determined by the apparent optical depth,as described in§3.Since the zero point of the system does not necessarily bear any relationship to a real kinematic component,we opt to present pixel velocity diﬀerences since they are independent of these considerations.In Figure9,we show the distribution of velocity diﬀerences for all pixel pairs in which both pixels have aﬂux between0.0and0.2,i.e.for the saturated pixels.In each panel, the model histogram is compared to both observed samples S1and S2(the full sample and one with possible double galaxies removed).The pure disk model distribution is quite narrow compared to the data,and the pure infall model is too broad.This is consistent with the visual impression from Figures4–8.As more disk component is added to a model the distribution gets narrower and the tail is reduced.Two–population models naturally have both the narrow distribution and the large tail at high velocity diﬀerences.For the0.0to0.2ﬂux bin,none of the models is in agreement with the Sample S1data according to a KS test.There are two reasons for this.First,the pixel by pixel statistics are quite sensitive tests.It is very diﬃcult to tune the parameters so that the distributions agree to the level where the KS test will not distinguish signiﬁcant diﬀerences.Second,there is a systematic diﬀerence between these models and the data which no amount ofﬁne tuning can alleviate.The diﬀerence is a deﬁcit of saturated pixel pairs in the bins with velocity diﬀerence20–80km s−1. The vertical velocity dispersion for the disk components in displayed models was V z=25km s−1. Increasing this value would reduce the inconsistency between the models and the data,however it would have to be increased substantially to bring them into agreement.At that point we would have a kinematic component that did not bear much resemblence to a thick disk.We conclude that there is a kinematic component contributing to the observed absorbers that is distinct from a thick disk or a infall/halo component.We also considered the velocity diﬀerence distributions for pixels withﬂux ranges0.2–0.4,。

kinematics(动力学)1动力学课件

The subjects that occupied physical scientists through the end of the nineteenth century----mechanics, light, heat, sound, electricity and magnetism-------are usually referred to as classical physics.
Velocity is the rate at which the position changes. • Average Velocity The average velocity of a particle during the time interval Δt is defined as the ratio of the displacement to the time interval.
It is classical physics we must master to understand the macroscopic world we live in, and classical physics is the main subject of the first part of the course.
K Δr and Δr
K Δr ≠ Δr
#mech001
Which of the following statements is (are) true ? (a) Δr is the magnitude of the displacement .
G (b) Δr is the magnitude of the displacement.
z
P1
G v1

chapter2_lecture

Genotype – pair of alleles present in an individual
Homozygous – two alleles of trait are the same (YY or yy)
Heterozygous – two alleles of trait are different (Yy)
Copyright © The McGraw-Hill Companies, Inc. Permission required to reproduce or display
4
Outline of Mendelian Genetics
The historical puzzle of inheritance and how Mendel’s experimental approach helped solve it
Mendel’s approach to genetic analysis including his experiments and related analytic tools
A comprehensive example of Mendelian inheritance in humans
First constructed true breeding lines for both traits, crossed them to produce dihybrid offspring, and examined the F2 for parental or recombinant types (new combinations not present in the parents)
Fig.2.9
2-9
Keys to Mendel’s experiments

chap1(kinematics)

限方向。加速度与速度的方向一般不同。
2. 自然坐标系
et ( t )
P1
s P 2 o
ds v vet et dt dv d(vet ) a dt dt dv det et v dt dt
et et (t t )
1 2 2 2 an v ( R ) R R
1
:
角加速度
2
* 运动学两类问题*
1. 已知：r r (t )，求： (t ), a (t ) v 2 dr d r 方法：求导数 v , a 2 dt dt 2. 已知： a (t ) 或 v (t ), 求： r (t )
et en
d det et lim en en lim dt dt t 0 t t 0 t
dv det dv dθ a et v et v en dt dt dt dt
d d ds 1 v dt ds dt
dv b v0t at 2 2 dt b t 1 2 b v0 y 2 2 2 b y v0
2
v y v0 v x by bv0t
2
2 0
2
2
2 0

END
b y v 2 bv0
Δ t 0 时，方向：Δ er // e er (t t ) er 大小：er er e er (t ) P er e der er lim lim e er (t ) dt t 0 t t 0 t P d e dt dr der dr d v er r er r e dt dt dt dt vr er v e

AP物理-谢涛-2讲义

lesson 2 Kinematics本节授课要点1 2 3Describing MotionUniform Acceleration MotionKinematics Problem Solving Strategy4567Graphical Analysis of Linear MotionFree Fall Motion and Projectile VerticallyVector Addition and Relative MotionProjectile Motiondescribing motions本节授课要点1 2 3Describing MotionUniform Acceleration MotionKinematics Problem Solving Strategy4567Graphical Analysis of Linear MotionFree Fall Motion and Projectile VerticallyVector Addition and Relative MotionProjectile Motionuniform acceleration motionKinema'c Equa'onsv=v 0+atx= x=v 0t+v 2=v 02+2ax 23412v 0+vt21kinematic problem solving skillsat 2本节授课要点1 2 3Describing MotionUniform Acceleration MotionKinematics Problem Solving Strategy4567Graphical Analysis of Linear MotionFree Fall Motion and Projectile VerticallyVector Addition and Relative MotionProjectile MotionKinema'c Equa'onsv=v 0+atx= x=v 0t+v 2=v 02+2ax 23412v 0+vt21kinematic problem solving skillsat 2always be aware that:four quantities: x, t, v, a ﬁve variables: x, t, v 0, v, a four equations: 4 variables in eachProblem Solving StepsStep1: Sketch a diagram to visually represent the problem. Step2: label the known variables or values on the diagram. Step3: Identify the unknown variable(s), and label them too. Step4: Choose the “star equation(s)”that contains all the 4 variables. Step5: Plug in the known values and solve for the unknown, or derive the unknown in terms of the known.Step 6: Feel proud in your mastery physics.kinematic problemsolving skills本节授课要点1 2 3Describing MotionUniform Acceleration MotionKinematics Problem Solving Strategy4567Graphical Analysis of Linear MotionFree Fall Motion and Projectile VerticallyVector Addition and Relative MotionProjectile Motiongraphical analysis of linear motion本节授课要点1 2 3Describing MotionUniform Acceleration MotionKinematics Problem Solving Strategy4567Graphical Analysis of Linear MotionFree Fall Motion and Projectile VerticallyVector Addition and Relative MotionProjectile Motionfree fall motion and projectile vertically F ree Fall: drop freelyfree fall motion and projectile vertically Projectile Vertically Up本节授课要点1 2 3Describing MotionUniform Acceleration MotionKinematics Problem Solving Strategy4567Graphical Analysis of Linear MotionFree Fall Motion and Projectile VerticallyVector Addition and Relative MotionProjectile Motionvector addition and relative motion vector additionParallelogram Rule and Parallelogram Methodvector addition and relative motion vector additionTail-to-tip Methodvector addition and relative motion vector additionAdding Vectors by Componentsvector addition and relative motion relative motion本节授课要点1 2 3Describing MotionUniform Acceleration MotionKinematics Problem Solving Strategy4567Graphical Analysis of Linear MotionFree Fall Motion and Projectile VerticallyVector Addition and Relative MotionProjectile Motionprojectile motion horizontally launched projectilesprojectile motion projectile launched at an angle回顾本节授课要点1 2 3Describing MotionUniform Acceleration MotionKinematics Problem Solving Strategy4567Graphical Analysis of Linear MotionFree Fall Motion and Projectile VerticallyVector Addition and Relative MotionProjectile Motion预告下节授课要点1 2 3ForcesNewton’s Laws of MotionNewton’s Laws Problem Solving Strategy Typical Problems of Applying Newton’s Laws4thank you。

kinematic Analysis

3Kinematic AnalysisThe kinematic constraint equations corresponding to the natural coordinates were explained in detail in Chapter 2, both for planar and three-dimensional multibody systems. They were then compared to other types of coordinates. Attention was also given to the main sources of constraint equations with natural coordinates: rigid body constraints, joint constraints, and the optional definition of relative or joint coordinates.In this chapter we will make use of those constraint equations to solve what is usually called kinematic problems, namely, the initial position or assembly problem, the finite displacement problem, and the velocity and acceleration analysis. The first two problems require an iterative solution of a system of nonlinear equations. Some special techniques to improve the convergence will be explained. Special attention will be addressed to the important case of over constrained multibody systems or, in general, to systems with non-independent constraint equations. The last section of this chapter is devoted to the case of non-holonomic joints.3.1 Initial Position ProblemThe initial position problem was defined in Section 1.3. It basically consists of determining the position of all the bodies in the system by knowing the positions of the fixed and the input bodies which can also be called guided or driven ele-ments. Mathematically, the initial position problem is reduced to determining from the known coordinates corresponding to the input elements the vector of dependent coordinates that satisfies the nonlinear system of constraint equa-tions. Note that the input can also be specified as the externally guided or driven linear or angular coordinates corresponding to several joints (as many joints as there are degrees of freedom) on which mixed coordinates have been defined. This basic notion is explained by means of the following examples:7172 3. Kinematic AnalysisFigure 3.1. Four-bar mechanism modeledwith natural coordinates.Figure 3.2. Four-bar mechanism modeled with mixed coordinates.Example 3.1As a first example, the four-bar mechanism of Figure 3.1 will be considered. This sys-tem has four natural coordinates (x 1,y 1,x 2,y 2).The constraint equations corresponding to this mechanism are following constant distance conditions:x 1 Ð x A 2 + y 1 Ð y A 2 Ð L 22 =02 Ð x 12 + y 2Ð y Ð L 32 = 0x 2 Ð x B 2 + y 2 Ð y B 2 Ð L 42 = 0These three equations are not sufficient to determine the four unknown variables of the problem. In fact, it is still necessary to enter the condition that the position of the input element (element 2) is known. If both coordinates of point 1 are known, then only two unknown variables are left. In this case, it is obvious that the first constraint equation which establishes the constant length condition of element 2 no longer makes any sense, because it does not contain any unknown variable. Consequently the prob-lem reduces to the finding of x 2 and y 2, using the last two nonlinear quadratic con-straint conditions.Example 3.2Let us consider the four-bar mechanism shown in Figure 3.2, which uses mixed coor-dinates; that is, the same coordinates as in example 3.1 plus the angle y between ele-ments 2 and 3 at joint 1. Let's assume that, instead of the position of the input element,one knows the angle y . In this case the constraint equations will be as follows (assuming a suitable value for y to be able to use the scalar product):x 1 Ð x A 2 + y 1 Ð y A 2 Ð L 22 = 02 Ð x 12 + y 2 Ð y Ð L 32 = 0x 2 Ð x B 2 + y 2 Ð y B 2 Ð L 42 = 0x 1 Ð x A 2 Ð x 1 +1 Ð y y 2 Ð y Ð L 2 L 3 cos y = 03.1 Initial Position Problem 73Figure 3.3. RSCR mechanism modeled with natural coordinates.which is a system with four equations and four unknown variables, assuming that the externally driven angle y is known.Example 3.3Figure 3.3 depicts a three-dimensional four-bar mechanism RSCR (Revolute-Spherical-Cylindrical-Revolute) modeled with natural coordinates. This mechanism has three movable points and one movable unit vector; that is, twelve dependent Cartesian coordinates and one degree of freedom. Also the input angle y has been in-troduced as an additional externally driven coordinate. The constraint equations corre-sponding to this mechanism are the following:1. 1 Ð x 0 Ð x y 1 Ð y A y 0 Ð y A 1 Ð z 0 Ð z k 1 cos y = 02.1 Ð x 1 Ð y 1 Ð z Ð k2 = 03.1 Ð x Ax 1 Ð y Ay 1 Ð z Az Ð k 3 = 04.x 2 Ð x 2 Ð y 12 + z 2 Ð z Ð k 4 = 05.x 2 Ð x 1x + y 2 Ð y 1y + z 2 Ð z 1z Ð k 5 = 0 6.u 1x 2 + u 1y 2 + u 1z 2 Ð 1= 07.x 3 - x B + y 3 - y B + z 3 - z B Ð k 6 = 08.x 3 Ð x B u 1x 3 Ð y 1y + z 3 Ð z B u 1z Ð k 7 = 09.x 3 Ð x B u Bx + y 3 Ð y B u By + z 3 Ð z B u Bz Ð k 8 = 010.u Bx u 1x + u B y u 1y + u B z u 1z Ð k 9 = 011.y 3 Ð y 1z 3 Ð z 2 u 1y = 012.z 3 Ð z 1x Ð x 3 Ð x 1z = 013.x 3 Ð x 1y Ð y 3 Ð y 1x = 0This is the system of nonlinear equations that governs the position problem for the RSCR mechanism. The first equation corresponds to the input angle definition; equa-74 3. Kinematic Analysistions 2 and 3 represent rigid body condition for element 2; equations 4 to 6 represent rigid body constraints for element 3; equations 6 to 10 represent the same for element 4, and equations 11 to 13 (only two of them are independent) contribute to define the cylindrical joint. Finally, k i (i=1,...,9) represents constant values.The above examples clearly indicate that irrespective of the multibody sys-tems being considered, the position problem is always based on solving the con-straint equations, which make up the following set of equations:F(q, t)=0(3.1) where q is the vector of the system dependent coordinates. It will be assumed that there are at least as many equations as there are unknown variables or coor-dinates. To solve systems of nonlinear equations such as (3.1), it is customary to resort to the well-known Newton-Raphson method which has quadratic conver-gence in the neighborhood of the solution (the error in each iteration is propor-tional to the square of the error in the previous iteration) and does not usually cause serious difficulties if one starts with a good initial approximation.The Newton-Raphson method is based on a linearization of the system (3.1) and consists in replacing this system of equations with the first two terms of itsexpansion in a Taylor series around a certain approximation qi to the desiredsolution. Once the substitution has been made, the system (3.1) becomesF(q, t)@F(q i)+F q(q i)(q Ð q i)=0(3.2) where the time variable has not been accounted for, so that in this problem has a constant value. Matrix F q is the Jacobian matrix for constraint equations; that is to say, the matrix of partial derivatives of these equations with respect to the de-pendent coordinates. This matrix takes the following form:F q=¶f1¶q1¶f1¶q2.......¶f1¶q n¶f2¶q1¶f2¶q2.......¶f2¶q n ...................¶f m¶q1¶f m¶q2.......¶f m¶q n(3.3)In equation (3.3), m is the number of constraint equations and n the number of dependent coordinates. If the constraint equations are independent, f=n-m is the number of degrees of freedom of the multibody system.Equation (3.2) represents a system of linear equations constituting an approx-imation to the nonlinear system (3.1). The vector q, obtained from the solution of equation (3.2), will be an approximation of the solution in (3.1). By calling this approximate solution qi+1, a recursive formula is obtained as follows:F(q i)+F q(q i)(q i+1Ð q i)=0(3.4)3.1 Initial Position Problem 75qFigure 3.4. Iteration process of the Newton-Raphson method.which can be used repeatedly until the error in the system of equations (3.1) is insignificant, or until the difference between the results of two successive itera-tions is smaller than a pre-specified tolerance. Figure 3.4 shows the geometric representation of the Newton-Raphson method for the case of a nonlinear equa-tion with one unknown. The function F(q) is linearized at point qi , i.e. substi-tuted by its tangent linear space, which are the first two terms of the Taylor ex-pansion formula. The point where the tangent space intersects the horizontal axisis the approximate solution qi+1. The function F(q) is again replaced at pointq i+1by the new tangent space and a new approximate solution qi+2is found. Onearrives ultimately within the desired accuracy to the exact solution q.Note that the Newton-Raphson iteration will not always converge to a solu-tion. It has been pointed out that if the initial approximation is not close enough to a solution, the algorithm may diverge. There is still another source of difficul-ties. If the values of the input variables do not correspond to a possible physical solution, the mathematical algorithm will fail irrespective of how the initial ap-proximation has been chosen.The Jacobian matrix of the constraint equations, defined by means of equa-tion (3.3), plays an extremely important role in all kinematic and dynamic analy-sis problems. In the equation (3.4), the Jacobian matrix determines the linear equation system used to find the successive approximations for solving the ini-tial position problem. Evaluating and triangularizing this matrix easily and quickly are characteristics of all good multibody system analysis methods. The natural coordinates permit the performance of these operations in the best possible way.In Section 1.2, it was stated that the initial position problem had multiple so-lutions, and this is indeed the case. Depending on the vector qo where the itera-tion begins, some solution will be attained.Example 3.4To complete this section on the initial position problem, the equations (3.4) corre-sponding to the four-bar mechanism studied in Examples 3.1 and 3.2 will be com-76 3. Kinematic Analysisa)b)Figure 3.5. Iteration process of the Newton-Raphson method in a four-bar mechanism.pletely developed. The constraint equations of this case were presented in Example3.1, and consequently the equation (3.4) takes the following form:2x1Ðx A y1Ðy A00x1Ðx2y1Ðy2x2Ðx1y2Ðy100x2Ðx B y2Ðy B ix1y1x2y2i+1Ðx1y1x2y2i==Ðx1Ðx+1ÐyÐL22x2Ðx+2Ðy12ÐL32x2Ðx+2ÐyÐL42iIn this system of equations, at least one of the four unknown coordinates must be known ahead of time in order to be able to solve the problem. If, for example, x1 is known, then:(x1)i+1 Ð (x1)i = 0and the first column of the Jacobian matrix is multiplied by zero, meaning that it can be eliminated.In the case of the four-bar mechanism of Figure 3.2, modeled with mixed coordi-nates and whose constraint equations are presented in Example 3.2, equation (3.4) be-comes:2(x1Ð x A)2(y1Ð y A)0002(x1Ð x2)2(y1Ð y2)2(x2Ð x1)2(y2Ð y1)0002(x2Ð x B)2(y2Ð y B)0 (x2Ðx1+x AÐx1)(y2Ðy1+y AÐy1)(x1Ð x A)(y1Ð y A)(L2 L3 sin y)i..x1y1x2y2y i+1Ðx1y1x2y2y i=(x1Ðx A)2+(y1Ðy A)2Ð L22(x2Ðx1)2+(y2Ðy1)2Ð L32(x2Ðx B)2+(y2Ðy B)2Ð L42(x1Ðx A)(x2Ðx1)+(y1Ðy A)(y2Ðy1)Ð L2 L3 cos y i3.1 Initial Position Problem 77Figure 3.6. Possible divergence in the Newton-Raphson iterationUsually, the angle y is known; therefore, the last unknown variable (y i+1 Ð y i ) has a zero value. Thus the fifth column of the Jacobian matrix can be eliminated.One characteristic common to the Jacobians matrices shown in this example (and in all the Jacobians matrices calculated with natural coordinates) is that they are linear functions of the dependent variables. For example, Figures 3.5a and 3.5b include drawings of the initial approximation, the first iterations, and the final solution of the initial position problem in the two four-bar mechanisms of Figures 3.1 and 3.2 com-puted according to the above expressions.The Newton-Raphson method, explained in this section, converges rather quickly (quadratic convergence) when it is close to the desired solution. At times, and during the first iterations, it can give very abrupt jumps as a result of not having started from a sufficiently good initial approximation. Figure 3.6shows what could happen in this case. The approximate solution q i+1 is further away from the true solution q , than the previous approximation q i . It is even possible that the value of function F (q ), a function that should be equal to zero,could increase when moving from q i to q i+1.This problem is not easy to solve without resorting to much more compli-cated numerical methods. In general, one should do everything possible to start from good initial approximations. If this cannot be achieved, then one should try to apply a reduction to the coordinates modification given by equation (3.4) and to apply it to the previous approximate solution q i . As this often works, a cor-rection factor of 1/2 or 1/3 is recommended. Finally, one should always make sure that the module of F (q ) decreases when going from point q i to q i+1.Some authors have solved the position problem at times by calculating differ-ent solutions numerically by means of the so-called continuation methods (Tsai and Morgan (1985)). Continuation methods start out from a position where the multibody system complies with all the constraint equations, although the input elements might not be at the desired position and the fixed joints might not be at78 3. Kinematic Analysistheir final position. With relaxed conditions regarding the input elements and the fixed element, it is not difficult to find a position on the multibody system that satisfies the constraint equations. Then, by means of small finite displacements whose convergence is guaranteed by the Newton-Raphson method, an attempt is made to move the input elements and the fixed joints to their correct position. At times, the bifurcation points (points at which two or more possible movements can occur) provide a way of finding different solutions to the position problem.3.2 Velocity and Acceleration Analysis3.2.1 Velocity AnalysisThe equations that permit solving the velocity problem originate after one dif-ferentiates with respect to time the constraint equations. If these equations are represented symbolically asF (q , t ) = 0(3.5)by differentiating with respect to time, the following equation is obtained:F ,q =ÐF t ºb (3.6)where F q is the Jacobian matrix defined by means of equation (3.3). Vector q is the vector of dependent velocities (derivative with respect to the time of the vec-tor of dependent coordinates or position variables). Vector (ÐF t = b ) is the par-tial derivative of the constraint equations with respect to time. If all the con-straints are scleronomous, meaning that there are no rheonomous or time depen-dent constraints, this derivative will be zero) If the position of the multibody system is known, equation (3.6) allows us to determine the velocities of the multibody system by starting from the velocity of the input elements. Just as in the position problem, the matrix that controls the velocity problem is the Jacobian matrix of the constraint equations. The essential difference between both problems is that where the position problem is nonlinear, the equations governing the velocity problem are linear. This means that the equations do not have to be iterated, and there is only one solution to a properly posed problem.The following example illustrates these concepts:Example 3.5As an example of this, the velocity equations of the four-bar mechanism of Figure 3.1will be determined below by using: a) relative coordinates, b) reference point coordi-nates, c) natural coordinates, and d) mixed coordinates.a)Using relative coordinates , the constraint equations are given by (See Section2.1.1),L 1 cos Y 1 + L 2 cos (Y 1 + Y 2) + L 3 cos (Y 1 + Y 2 + Y 3) Ð O D = 03.2 Velocity and Acceleration Analysis 79L1 sin Y1+ L2 sin (Y1+Y2)+ L3 sin (Y1+Y2+Y3)=0Differentiating these equations with respect to time, we obtain:ÐL1 sin y1y1Ð L2 sin (y1+y2)(y1+y2)ÐÐ L3 sin (y1+y2+y3)(y1+y2+y3)=0L1 cos y1y1+ L2 cos (y1+y2)(y1+y2)++ L3 cos (y1+y2+y3)(y1+y2+y3)=0and by rearranging these equations, we arrive at:Ð L1s1Ð L2s12Ð L3s123Ð L2s12Ð L3s123Ð L3s123 L1c1+ L2c12+ L3c123 L2c12+ L3c123 L3c123y1 y2 y3where s1 = sin y1, s12 = sin (y1 + y2), and so forth.If one of the three velocities in the previous equation is known such as the one cor-responding to the input coordinate) the corresponding column of the Jacobian matrix can be moved to the right-hand side of the equation. This results in a system of two linear equations with two unknown velocities that can be solved with no difficulties.b)Using reference point coordinates,the constraint equations are represented by (SeeSection 2.1.2):x1ÐxÐL12 cos Y1=y1ÐyÐL12 sinY1=02Ðx1ÐL12 cos Y1Ð L22 cos Y2=y2ÐyÐL12 sin Y1Ð L22 sinY2=03Ðx2ÐL22 cos Y2Ð L32 cos Y3=y3ÐyÐL22 sin Y2Ð L32 sin Y3=0x3Ðx DÐL32 cos Y3=0y3Ðy DÐL32 sin Y3=0and the time derivatives are:x1+ L1/2Y1 sin Y1=0y1Ð L1/2Y1 cos Y1=0x2Ð x1+L12Y1 sin Y1+L22Y2 sin Y2=0y2Ð y1ÐL12Y1 cos Y1ÐL22Y2 cos Y2=0x3Ð x2+L22Y2 sin Y2+L32Y3 sin Y3=0y3Ð y2ÐL22Y2 cos Y2ÐL32Y3 cos Y3=0x3+L32y3 sin Y3=0y3ÐL32Y3 cos Y3=080 3. Kinematic AnalysisThese equations can be expressed in matrix form as follows:F q q =0where the matrix F q is10s1L1/200000001Ðc1L1/2000000Ð10s1L1/210s2L2/20000Ð1Ðc1L1/201Ðc2L2/2000000Ð10s2L2/210s3L3/20000Ð1Ðc2L2/201Ðc3L3/200000010s3L3/200000001Ðc3L3/2 which is a system of eight equations with nine unknown velocities. If the angular ve-locity y1 is known for element 2, the third column of the Jacobian matrix will be moved to the right side member. The result will be a system of eight linear equations with eight unknown velocities.c)With natural coordinates the constraint equations (Section 2.1.3) are representedby:(x1Ð x A)2+(y1Ð y A)2Ð L22=0(x2Ð x1)2+(y2Ð y1)2Ð L32=0(x3Ð x B)2+(y3Ð y B)2Ð L42=0whose time derivatives are:(x1Ð x A) x1+(y1Ð y A) y1=0(x2Ð x1)(x2Ð x1)+(y2Ð y1)(y2Ð y1)=0(x2Ð x B) x2+(y2Ð y B) y2=0and in matrix form yields:(x1Ðx A)(y1Ðy A)00 (x1Ðx2)(y1Ðy2)(x2Ðx1)(y2Ðy1) 00(x2Ðx B)(y2Ðy B)x1 y1 x2 y2By knowing one of the four natural velocities and by moving the corresponding column to the right-hand side of this equation, one can find the remaining velocities with the resulting set of three linear equations and three unknown variables.d)Using mixed coordinates, the constraint equations (Section 2.1.4) are:(x1Ð x A)2+(y1Ð y A)2Ð L22=0(x2Ð x1)2+(y2Ð y1)2Ð L32=0(x3Ð x B)2+(y3Ð y B)2Ð L42=0(x1Ð x A)(x2Ð x1)+(y1Ð y A)(y2Ð y1)Ð L2 L3 cos Y=03.2 Velocity and Acceleration Analysis 81Figure 3.7. Results of a velocity analysis in a four-bar mechanism.differentiating with respect to time:x1Ðx A x1+y1Ðy A y1=0x2Ðx x2Ðx+y2Ðy y2Ðy=0x2Ðx B x2+y2Ðy B y2=02Ðx1 x1+x1Ðx AÐx1+2Ðy1 y1++y1Ðy A y2Ðy1+L2L3 sin y y=0which can be expressed in matrix form as1Ðx1Ðy0001Ðx21Ðy22Ðx12Ðy10002Ðx2Ðy0x2Ð2x1+x A y2Ð2y1+y A1Ðx1Ðy L2L3sin yx1y1x2y2y=If y is known, the fifth column will be moved to the right-hand side and will leave four equations with four unknowns. Figure 3.7 shows the result of a velocity analysis in accordance with an input velocity of y =1.3.2.2 Acceleration AnalysisThe finding of the dependent acceleration vector q becomes apparent by simply differentiating with respect to time the velocity equation (3.6). This yields the following result:F q,q=ÐtÐq q º c(3.7)If the position vector q and the velocity vector q are known, by solving the system of linear equations (3.7), one can find the dependent acceleration vector q. Note that the leading matrix of the systems of linear equations (3.6) and (3.7) is exactly the same. As a consequence, if it has been formed and triangularized82 3. Kinematic AnalysisFigure 3.8. Results of an acceleration analysis in a four-bar mechanism.to solve the velocity problem, the acceleration analysis can be carried out by simply forming the right-hand side and by performing a forward reduction and backward substitution. When there are no rheonomous or time-dependent con-straints, the velocity problem is homogeneous; whereas the acceleration problem is always non-homogeneous as long as the velocities are not equal to zero.Equation (3.7) can be differentiated once again to obtain the jerk or over ac-celeration equation:F q d dt = Ð t Ð 2 q q Ð q q (3.8)Once again a system of linear equations has been obtained whose leadingmatrix is the Jacobian matrix of the constraint equations.Example 3.6Included below are the acceleration equations for the four-bar mechanism of Example3.5d modeled with mixed coordinates. These equations are obtained by differentiating the corresponding velocity equations:1Ðx 1Ðy 0001Ðx 21Ðy 22Ðx 12Ðy 10002Ðx 2Ðy 0x 2Ð2x 1+x A y 2Ð2y 1+y A1Ðx 1Ðy L 2 L 3 sin y = Ð x 1y 1000x 1Ðx y 1Ðy x 2Ðx y 2Ðy 000x 2y 20x 2Ð2x 1y 2Ð21x 1y 1L 2 L 3 cos y x 1y 1x 2y 2y3.2 Velocity and Acceleration Analysis 83If all the velocities q and input accelerations y are known, the remaining accelera-tions q can be calculated by means of the four equations with four unknowns resulting from moving the fifth column, multiplied by y, to the RHS of the acceleration equa-tion. Figure 3.8 graphically shows the result of an acceleration analysis that corre-sponds to the expression developed in this example.3.3 Finite Displacement AnalysisThe finite displacement analysis is closely related to the initial position problem, and is controlled by the same system of nonlinear equations (the kinematic constraint equations). The velocity and acceleration analyses are used at times in finite displacement analysis to improve the initial approximation with which the iterative process begin, which explains the reason for including it here and not immediately after the initial position problem.3.3.1 Newton-Raphson IterationAs explained in Section 1.2, once one knows a position of the multibody system where all the constraint equations are satisfied, the finite displacement problem consists of finding the new position that the system takes when a finite displace-ment is applied to each one of the input elements or externally driven relative coordinates. Finite displacement is understood to be any movement other than infinitesimal.The main problem dealt with in this section is of the same nature and conse-quently controlled by the same equations of the position problem. Therefore, the Newton-Raphson method can be used for solving it. The difference between both problems lies in the fact that the finite displacement problem usually relies on a good initial approximation which is obtained from a previous exact posi-tion where all the elements satisfy the constraints. It is possible to improve upon the approximation by means of a velocity and acceleration analysis, as will be described in the next section.These advantages do away with many of the convergence problems encoun-tered in the initial position problem. In addition, the problem of multiple solu-tions becomes marginal. If the displacement of the input elements is small enough, then of all the possible solutions for the constraint equations, the correct one will be the closest to the starting position. This is precisely the one obtained by the Newton-Raphson iterations. However, there still remains the possibility of driving or trying to drive the multibody system to unfeasible positions, that is, positions that cannot be reached without violating some constraints equations. Trying to move the end-effector of a robot out of its workspace is an example of a finite displacement problem where the Newton-Raphson method will necessar-ily fail to find a correct solution.84 3. Kinematic AnalysisBFigure 3.9. Improving the initial approximation.A2' BFigure 3.10. Better estimate for an initial approximation.3.3.2 Improved Initial ApproximationIn order to determine and improve the initial or starting approximation, the ex-ample of the four-bar mechanism will be used once again. This will clearly de-scribe the method without any loss of generality.Figure 3.9 shows a four-bar mechanism, in which the input element has been rotated a finite angle. One possible way of generating an initial approximation is by not varying the remaining natural coordinates as in the starting position shown in Figure 3.9. This approximation leads to a severe violation of the con-straint equations.The initial approximation shown in Figure 3.9 can be improved upon by means of velocity analysis, as indicated in Figure 3.10. The velocity analysis is carried out by imposing a velocity at the input element so that the endpoint 1' of the velocity vector of 1 is the closest point to 1" over the perpendicular to A-1 (1'-1" is parallel to A-1).3.3 Finite Displacement Analysis 85Since 1" is known, it is not difficult to determine the velocity of the input el-ement such that the end of the velocity vector at point 1 is 1'. Using this velocity as input, a velocity analysis is performed, and the ends of the velocity vectors are determined for all the basic points of the mechanism (in Figure 3.10, 2' is the end of the velocity vector of point 2).The initial approximation used to start the iterations for the Newton-Raphson method is indicated by the dotted lines in Figure 3.10 It is an improvement over the one in Figure 3.9. Note that the initial approximation is (A-1"-2'-B) and not (A-1'-2'-B). The exact position 1" of point 1 is known because it belongs to the input element and this exact position should be used.It is not essential that point 1' be the closest one to 1" on the tangent to the trajectory of 1. Another simpler possibility for calculating point 1' and the velocity of the input element, is to assume that point 1 changes to position 1" in an arbitrary period of time such as 1 second. Next, calculate the angular velocity of the input element by dividing angle 1-1" (in radians) by the said amount of time where the quotient is the said angular velocity. The position at the initial approximation of any point P can be calculated by means of the following expression:q = q o + q D t (3.9) Equation (3.9) is an approximate integral of velocities starting from the previ-ous position. An approximate integration which also causes the accelerations to intervene can be obtained in a similar manner:q D t2(3.10)q = q o + q D t + 12This formula suggests that the initial approximation can be constructed start-ing from a velocity analysis and an acceleration analysis. To calculate the veloc-ity and acceleration of the input element one may proceed as follows:1.Apply one of the previously studied methods and determine the velocity ofthe input elements.2.Knowing the initial and final position of the input elements and their veloc-ity, determine the acceleration to be applied to them applying equation (3.10) to the input elementsDetermination of the initial approximation by means of velocity and accelera-tion analysis allows the iterations to begin with a better approximation to the fi-nal solution. The cost of an acceleration analysis is small if a velocity analysis has already been performed. The matrix for both systems of equations is the same, and one only needs to form it and triangularize it once. Based on the expe-rience gained through numerical experiments performed by the authors, the ini-tial approximation constructed with velocities and accelerations does not always give better results than the one determined from velocities only.。

专业英语Kinematics

an

2
nˆ
Where is the radius of curvature.
Quantum Physics and Wave Mechanics
The Theory of Relativity
Part One Mechanics
Chapter 1 Kinematics —Uniformly Accelerated Motion
Kinematics is the study of the geometry of motion: it deals with the mathematical description of motion in terms of position, velocity, and acceleration. Kinematics serves as a prelude to dynamics, which studies force as the cause of
numbers, depending on . Because these quantities
have signs, and there fore implied directions along
known axes, it is common to refer to them as
velocities. The magnitude of velocity is given
Velocity is a vector quantity. If an object undergoes
a vector displacement r in a time interval t ,then
r (t t) r (t) r

foundamental

Fundamental 1 of 2 adjective基本 1 的2形容词1a:serving as a basis supporting existence or determining essential structure or function:BASIC作为支撑存在或确定基本结构或功能的基础：基本Responsibility is fundamental to democracy.责任是民主的根本。

The Constitution ensures our fundamental rights.《宪法》保障我们的基本权利。

b:serving as an original or generating source:PRIMARY用作原始或生成源：主a discovery fundamental to modern computers现代计算机的基础发现2a:of or relating to essential structure, function, or facts:RADICAL 的或与基本结构、功能或事实有关：部首Fundamental change根本性变化also:of or dealing with general principles rather than practical application另外：或处理一般原则而不是实际应用fundamental science基础科学b: adhering to FUNDAMENTALISM坚持原教旨主义a preacher who is evangelical, Bible-teaching, and fundamental 一个福音派、圣经教导和基本派的传道人3:of central importance:PRINCIPAL中心重要性：Fundamental purpose基本目的Such fundamental events as birth, marriage, and death诸如出生、结婚和死亡等基本事件4:belonging to one's innate or ingrainedcharacteristics:DEEP-ROOTED属于一个人的先天或根深蒂固的特征：根深蒂固Her fundamental good humor她基本的幽默5:of, relating to, or produced by the lowest component of a complex vibration的、与复数振动的最低分量有关或由其产生Fundamental 2 of 2noun基本 2 的2名词1:something fundamental一些基本的东西especially:one of the minimum constituents without which a thing or a system would not be what it is特别是：一个最小成分之一，没有它，一个事物或一个系统就不会是现在的样子2a:the principal musical tone produced by vibration (as of a string or column of air) on which a series of higher harmonics is based 由振动（如一根弦或一列空气）产生的主要音乐音调，一系列高次谐波的基础是振动b:the root of a chord和弦的根部3:the harmonic component of a complex wave that has the lowest frequency and commonly the greatest amplitude频率最低，通常振幅最大的复杂波的谐波分量。

Princeton lectures 2

Cchapter 2KinematicsDisplacementս〫Displacement, distanceս〫ˈ䐍⿫Speed and Velocity䙏⦷઼䙏ᓖVelocity, average velocity, average speed䙏ᓖˈᒣ൷䙏ᓖˈᒣ൷䙏⦷average speed =average velocity =Acceleration࣐䙏ᓖAcceleration, average acceleration࣐䙏ᓖˈᒣ൷࣐䙏ᓖaccelerationर࣐䙏ᓖUniformaverage acceleration =Uniformly Accelerated Motion and the Big Fiveर࣐䙏ⴤ㓯䘀ࣘFree Fall㠚⭡㩭փGravitational accelerationзᴹᕅ࣐࣋䙏ᓖ˄䟽࣐࣋䙏ᓖ˅Projectile Motionᣋփ䘀ࣘLaunch angle, range, and parabolic pathࠪሴ䀂ˈ㤳തˈᣋ⢙㓯Kinematics with Graphs䘀ࣘᆖമۿPosition versus time, velocity versus time, slope, area, curvature ս〫ᰦ䰤മۿˈ䙏ᓖᰦ䰤മۿˈᯌ⦷ˈ䶒〟ˈᴢ⦷The slope of a position-versus-time graph gives the velocity.The slope of a velocity-versus-time graph gives the acceleration.Given a velocity-versus-time graph, the area between the graph and the t-axis is equal to the object’s displacement.Chapter 2 Review Questions1. A rock is thrown straight upward from the edge of a 30 m cliff, rising 10 m then falling all the way down to the base of the cliff. Find the rock’s displacement.2. In a track-and-field event, an athlete runs exactly once around an oval track, a total distance of 500 m. Find the runner’s displacement for the race.3. Assume that the runner in sample question 3 completes the race in 1 minute and 20 seconds. Find her average speed and the magnitude of her average velocity.4. Is it possible to move with constant speed but not constant velocity? Is it possible to move with constant velocity but not constant speed?5.A car is traveling in a straight line along a highway at a constant speed of 80 miles per hour for 10 seconds. Find its acceleration.6. A car is traveling in a straight line along a highway at a speed of 20 m/s. The driver steps on the gas pedal and, 3 seconds later, the car’s speed is 32 m/s. Find its average acceleration.7. Spotting a police car ahead, the driver of the car in the previous example slows from 32 m/s to 20 m/s in 2 sec.Find the car’s average acceleration.8. An object with an initial velocity of 4 m/s moves along a straight axis under constant acceleration. Three seconds later, its velocity is 14 m/s. How far did it travel during this time?9. A car that’s initially traveling at 10 m/s accelerates uniformly for 4 seconds at a rate of 2 m/s 2in a straight line. How far does the car travel during this time?10. A rock is dropped off a cliff that’s 80 m high. If it strikes the ground with an impact velocity of 40 m/s, what acceleration did it experience during its descent?Questions 11-12The velocity of an object as a function of time is given by the following graph:11. At which point (A,B,C,D, or E) is the magnitude of the acceleration the greatest?12. How would you answer this same question if the graph shown were a position-versus-time graph?13. A rock is dropped from an 80-meter cliff. How long does it take to reach the ground?14.One second after being thrown straight down, an object is falling with a speed of 20 m/s. How fast will it be falling 2 seconds later?15. If an object is thrown straight upward with an initial speed of 8 m/s and takes 3 seconds to strike the ground, from what height was the object thrown?16. An object is thrown horizontally with an initial speed of 10 m/s. How far will it drop in 4 seconds?17. From a height of 100 m, a ball is thrown horizontally with an initial speed of 15 m/s. How far GRHV LW WUDYHO KRUL]RQWDOO\ LQ WKH ILUVW ௗVHFRQGV"18. A projectile is traveling in a parabolic path for a total of 6 seconds. How does its horizontal velocity 1 s after launch compare to its horizontal velocity 4 s after launch?19. An object is projected upward with a 30° launch angle and an initial speed of 60 m/s. How many seconds will it be in the air? How far will it travel horizontally?1. An object that’s moving with constant speed travels once around a circular path. True statements about this motion include which of the following?I. The displacement is zero.II. The average speed is zero.III. The acceleration is zero.(A) I only(B) I and II only(C) I and III only(D) III only(E) II and III only2. At time t=t1, an object’s velocity is given by the vector v1shown below.A short time later, at t=t2, the object’s velocity is the vector v2.If v1and v2have the same magnitude, which one of the following vectors best illustrates the object’s average acceleration between t=t1and t=t2?(A)(B)(C)(D)(E)3. Which of the following must always be true?I. If an object’s acceleration is constant, then it must move in a straight line.II. If an object’s acceleration is zero, then its speed must remain constant.III. If an object’s speed remains constant, then its acceleration must be zero.(A) I and II only (B) I and III only (C) II only(D) III only(E) II and III only4. A baseball is thrown straight upward. What is the ball’s acceleration at its highest point?(A) 0(B)g , downward(C)g , downward(D)g , upward(E)g , upward5. How long would it take a car, starting from rest and accelerating uniformly in a straight line at 5 m/s 2, to cover a distance of 200 m ?(A) 9.0 s(B) 10.5 s(C) 12.0 s(D) 15.5 s(E) 20.0 s6. A rock is dropped off a cliff and strikes the ground with an impact velocity of 30 m/s. How high was the cliff?(A) 15 m(B) 20 m(C) 30 m(D) 45 m(E) 60 m7.A soccer ball, at rest on the ground, is kicked with an initial velocity of 10 m/s at a launch angle of 30°. Calculate its total flight time, assuming that air resistance is negligible.(A)0.5 s(B) 1 s(C) 1.7 s(D) 2 s(E) 4 s8. A stone is thrown horizontally with an initial speed of 30 m/s from a bridge. Find the stone’s total speed when it enters the water 4 seconds later. (Ignore air resistance.)(A)30 m/s(B) 40 m/s(C) 50 m/s(D) 60 m/s(E) 70 m/s9. Which one of the following statements is true concerning the motion of an ideal projectile launched at an angle of 45° to the horizontal?(A) The acceleration vector points opposite to the velocity vector on the way up and in the same direction as the velocity vector on the way down.(B) The speed at the top of the trajectory is zero.(C) The object’s total speed remains constant during the entire flight.(D) The horizontal speed decreases on the way up and increases on the way down.(E) The vertical speed decreases on the way up and increases on the way down.。

基于自转一阶非连续式微球双平盘研磨的运动学分析与实验研究

第53卷第8期表面技术2024年4月SURFACE TECHNOLOGY·133·基于自转一阶非连续式微球双平盘研磨的运动学分析与实验研究吕迅1,2*，李媛媛1，欧阳洋1，焦荣辉1，王君1，杨雨泽1（1.浙江工业大学机械工程学院，杭州 310023；2.新昌浙江工业大学科学技术研究院，浙江绍兴 312500）摘要：目的分析不同研磨压力、下研磨盘转速、保持架偏心距和固着磨料粒度对微球精度的影响，确定自转一阶非连续式双平面研磨方式在加工GCr15轴承钢球时的最优研磨参数，提高微球的形状精度和表面质量。

方法首先对自转一阶非连续式双平盘研磨方式微球进行运动学分析，引入滑动比衡量微球在不同摩擦因数区域的运动状态，建立自转一阶非连续式双平盘研磨方式下的微球轨迹仿真模型，利用MATLAB对研磨轨迹进行仿真，分析滑动比对研磨轨迹包络情况的影响。

搭建自转一阶非连续式微球双平面研磨方式的实验平台，采用单因素实验分析主要研磨参数对微球精度的影响，得到考虑圆度和表面粗糙度的最优参数组合。

结果实验结果表明，在研磨压力为0.10 N、下研磨盘转速为20 r/min、保持架偏心距为90 mm、固着磨料粒度为3000目时，微球圆度由研磨前的1.14 μm下降至0.25 μm，表面粗糙度由0.129 1 μm下降至0.029 0 μm。

结论在自转一阶非连续式微球双平盘研磨方式下，微球自转轴方位角发生突变，使研磨轨迹全覆盖在球坯表面。

随着研磨压力、下研磨盘转速、保持架偏心距的增大，微球圆度和表面粗糙度呈现先降低后升高的趋势。

随着研磨压力与下研磨盘转速的增大，材料去除速率不断增大，随着保持架偏心距的增大，材料去除速率降低。

随着固着磨料粒度的减小，微球的圆度和表面粗糙度降低，材料去除速率降低。

关键词：自转一阶非连续；双平盘研磨；微球；运动学分析；研磨轨迹；研磨参数中图分类号：TG356.28 文献标志码：A 文章编号：1001-3660(2024)08-0133-12DOI：10.16490/ki.issn.1001-3660.2024.08.012Kinematic Analysis and Experimental Study of Microsphere Double-plane Lapping Based on Rotation Function First-order DiscontinuityLYU Xun1,2*, LI Yuanyuan1, OU Yangyang1, JIAO Ronghui1, WANG Jun1, YANG Yuze1(1. College of Mechanical Engineering, Zhejiang University of Technology, Hangzhou 310023, China;2. Xinchang Research Institute of Zhejiang University of Technology, Zhejiang Shaoxing 312500, China)ABSTRACT: Microspheres are critical components of precision machinery such as miniature bearings and lead screws. Their surface quality, roundness, and batch consistency have a crucial impact on the quality and lifespan of mechanical parts. Due to收稿日期：2023-07-28；修订日期：2023-09-26Received：2023-07-28；Revised：2023-09-26基金项目：国家自然科学基金（51975531）Fund：National Natural Science Foundation of China (51975531)引文格式：吕迅, 李媛媛, 欧阳洋, 等. 基于自转一阶非连续式微球双平盘研磨的运动学分析与实验研究[J]. 表面技术, 2024, 53(8): 133-144.LYU Xun, LI Yuanyuan, OU Yangyang, et al. Kinematic Analysis and Experimental Study of Microsphere Double-plane Lapping Based on Rotation Function First-order Discontinuity[J]. Surface Technology, 2024, 53(8): 133-144.*通信作者（Corresponding author）·134·表面技术 2024年4月their small size and light weight, existing ball processing methods are used to achieve high-precision machining of microspheres. Traditional concentric spherical lapping methods, with three sets of circular ring trajectories, result in poor lapping accuracy. To achieve efficient and high-precision processing of microspheres, the work aims to propose a method based on the first-order discontinuity of rotation for double-plane lapping of microspheres. Firstly, the principle of the first-order discontinuity of rotation for double-plane lapping of microspheres was analyzed, and it was found that the movement of the microsphere changed when it was in different regions of the upper variable friction plate, resulting in a sudden change in the microsphere's rotational axis azimuth and expanding the lapping trajectory. Next, the movement of the microsphere in the first-order discontinuity of rotation for double-plane lapping method was analyzed, and the sliding ratio was introduced to measure the motion state of the microsphere in different friction coefficient regions. It was observed that the sliding ratio of the microsphere varied in different friction coefficient regions. As a result, when the microsphere passed through the transition area between the large and small friction regions of the upper variable friction plate, the sliding ratio changed, causing a sudden change in the microsphere's rotational axis azimuth and expanding the lapping trajectory. The lapping trajectory under different sliding ratios was simulated by MATLAB, and the results showed that with the increase in simulation time, the first-order discontinuity of rotation for double-plane lapping method could achieve full coverage of the microsphere's lapping trajectory, making it more suitable for precision machining of microspheres. Finally, based on the above research, an experimental platform for the first-order discontinuity of rotation for double-plane lapping of microsphere was constructed. With 1 mm diameter bearing steel balls as the processing object, single-factor experiments were conducted to study the effects of lapping pressure, lower plate speed, eccentricity of the holding frame, and grit size of fixed abrasives on microsphere roundness, surface roughness, and material removal rate. The experimental results showed that under the first-order discontinuity of rotation for double-plane lapping, the microsphere's rotational axis azimuth underwent a sudden change, leading to full coverage of the lapping trajectory on the microsphere's surface. Under the lapping pressure of 0.10 N, the lower plate speed of 20 r/min, the eccentricity of the holder of 90 mm, and the grit size of fixed abrasives of 3000 meshes, the roundness of the microsphere decreased from 1.14 μm before lapping to 0.25 μm, and the surface roughness decreased from 0.129 1 μm to 0.029 0 μm. As the lapping pressure and lower plate speed increased, the microsphere roundness and surface roughness were firstly improved and then deteriorated, while the material removal rate continuously increased. As the eccentricity of the holding frame increased, the roundness was firstly improved and then deteriorated, while the material removal rate decreased. As the grit size of fixed abrasives decreased, the microsphere's roundness and surface roughness were improved, and the material removal rate decreased. Through the experiments, the optimal parameter combination considering roundness and surface roughness is obtained: lapping pressure of 0.10 N/ball, lower plate speed of 20 r/min, eccentricity of the holder of 90 mm, and grit size of fixed abrasives of 3000 meshes.KEY WORDS: rotation function first-order discontinuity; double-plane lapping; microsphere; kinematic analysis; lapping trajectory; lapping parameters随着机械产品朝着轻量化、微型化的方向发展，微型电机、仪器仪表等多种工业产品对微型轴承的需求大量增加。

科技英语力学部分

Rectilinear motion 直线运动
curvilinear motion 曲线运动 Circular motion 圆周运动 Uniform circular motion 匀速圆周运动
词汇
frame of reference 参考系；
coordinate system 坐标系；
Cartesian coordinate system 直角坐标系; the nature coordinate system 自然坐标系； Particle 质点； position vector 位置矢量（位矢）； Velocity 速度;
Mechanics 力学；Kinematics 运动学；dynamics 动力学
Theoretical Mechanics 理论力学
Quantum Mechanics 量子力学 Fluid Mechanics 流体力学 Computational Mechanics 计算力学 Thermodynamics 热力学 Mechanical motion 机械运动
FRAME OF REFERENCE

The fact that if we choose different body as the reference to describe the same motion of a given object, the indications will be different, is called the relativity of the description and measurement of the motion. The particular body that serves as a reference chosen by the observer is defined as the frame of reference.

mankiw7e-chap02

G includes all government spending on goods
and services.
G excludes transfer payments
(e.g., unemployment insurance payments), because they do not represent spending on goods and services.
total output
total expenditure
the sum of value-added at all stages
in the production of final goods
CHAPTER 2
The Data of Macroeconomics
19
Real vs. nominal GDP
CHAPTER 2
The Data of Macroeconomics
14
FYI: U.S. Government Spending, 2008
$ billions
Govt spending $ቤተ መጻሕፍቲ ባይዱ,882.4
% of GDP
20.2%
- Federal
Non-defense
1,071.9
337.0
Unsold output goes into inventory,
and is counted as “inventory investment”… …whether or not the inventory buildup was intentional.
In effect, we are assuming that
Chapter 2: The Data of Macroeconomics

L2_Kinematics_and_Dynamics

1
Kinematics Definition
• Kinematics: The study of motion in space • Assume:
(without regard to the forces which cause it)
A
B
• Points with right-hand Frames • Rigid-bodies in 3D-space (6-dof) • 1-dof joints: Rotary (R) or Prismatic (P) (5 constraints)
• There isn t a unique method of specifying the angles
that define these transformations
Slide 14
MTRX4700 Experimental Robotics Kinematics
7
Position and Orientation [7]
The orientation of frame {B} relative to coordinate frame {A}
• Columns are {B} written in {A}
Slide 11
MTRX4700 Experimental Robotics Kinematics
Position and Orientation [3]
• • • • • • steering (control, simulation) perception (image formation) sensor head and communication antenna pointing world modelling (maps, object models) terrain following (control feedforward) gait control of legged vehicles

Chapter 2 kinematics

Chapter 2 Kinematics2-1. Air resistance acting on a falling body can be taken by the approximate relation for the acceleration: a g kv=- where k is a constant. Derive a formula for the velocity of the body as a function of time assuming it stars from rest (v=0 at t=0).2-2. The acceleration of a particle is given by a=t=0, v=10m/s and x=0 (a) What is the speed as a function of time? (b) What is the displacement as a function of time? (c) What are the acceleration, speed and displacement at t=5.0s?2-3. The position of a particle as a function of time is given by()2=+-. Determine the particle’s velocity and acceleration7.608.85r ti j t k mas a function of time.2-4. At t=0, a particle stars from rest and moves in the xy plane with an acceleration ()2=+. Determine (a) the x and y components ofa i j m s4.0 3.0/velocity. (b) The speed of the particle, and (c) the position of the particle, all as a function of time.2-5. The position of a particle moving in the xy plane is given by()=+, where r is in meter and t is in second. (a) Show that r t i t j2cos32sin3this represents circular motion of radius 2m centered at the origin. (b) Determine the velocity and acceleration vectors as function of time. (c) Determine the speed and magnitude of the acceleration. (d) Show that the acceleration vector always points toward the center of the circle.2-6. A swimmer is capable of swimming 1.00m/s in still water. (a) If she aims her body directly across a 75m wide river whose current is 0.80m/s, how far downstream (from a point opposite her starting point) will she land? (b) How long will it take her to reach the other side?Ans ：2-1 (1)--kt g e k 2-2 (a) 3/2410(/)3=+v t m s (b) 5/2810()15=+x t t m (c) 24.5 /25 /80 .===am s v m s x m ，， 2-3 7.602=-v i tk ， 2=-a k 2-4 (a) 4=x v t 3=y v t (b) 5t (c) 222.0 1.5()=+r t i t j m2-5 (b) 6.0sin3.0 6.0cos3.0=-+v ti tj ， 18.0sin3.018.0cos3.0=--a ti tj (c) 6.0/=v m s 218.0/=a m s (d) Hint: 9.0a r =-2-6 (a) 60m (b)75s。

3 - Kinematics in One Dimension

This idealized motion is called free-fall and the acceleration of a freely falling body is called the acceleration due to gravity.
g 9.80 m s
2013年7月22日
2.4 Equations of Kinematics for Constant Acceleration
x vot at
1 2
2 1 2
ቤተ መጻሕፍቲ ባይዱ6.0 m s 8.0 s 110 m
2013年7月22日
2.0 m s 8.0 s
2
2
19
2.4 Equations of Kinematics for Constant Acceleration
2
or
32.2 ft s
2
25
2.6 Freely Falling Bodies
g 9.80 m s
2
2013年7月22日
26
2.6 Freely Falling Bodies
Example 10 A Falling Stone A stone is dropped from the top of a tall building. After 3.00s of free fall, what is the displacement y of the stone?
DEFINITION OF AVERAGE ACCELERATION
v v o v a t to t
2013年7月22日
13
2.3 Acceleration

SAT2物理习题 Kinematics

SAT2 物理习题Kinematics以下是小编为大家整理的SAT2 物理习题Kinematics，希望对大家有帮助，跟随SAT 资料下载的小编一起来练习练习吧!1.An object that’s moving with constant speed travels once around a circular path. Which of the followingis/are true concerning this motion?I.The displacement is zero.II.The average speed is zero.III. The acceleration is zero.(A) I only(B) I and II only(C) I and III only(D) III only(E) II and III only2. At time t = t1 ，an object’s velocity is given by the vector v1, shown below.A short time later, at t = t2, the object’s velocity is the vector v2.If v1 and v2 have the same magnitude, which one of the following vectors best illustrates tfae object’s average acceleration between t=t1and t=t2?3. Which of the following is/are true?I. If an object’s accelCTation is constant, then it must move in a straight line.II.If an object’s acceleration is zero,then its speed must remain constant.III.If an object’s speed remains constant,then its acceleration must be zero.(A) I and II only(B) I and III only(C) II only(D) III only(E) II and III only4. A baseball is thrown straight upward. What is the ball’s acceieration at its hipest po int?(A) 0(B) 1/2g,downward(C) g,downward(D) 1/2g,upward(E) g,upward5. How long would it take a car, starting from rest and accelerating uniformly in a straight line at 5m/s2, to cover a distance of200m?(A) 9.0 s(B) 10.5 s(C)12.0s(D) 15.5 s(E) 20.0 s6. A rock is dropped off a cliff and strikes the ground with an impact velocity of 30 m/s. How high was the cliff?(A) 15 m(B) 20 m(C) 30 m(D) 45m(E) 60 m7. A stone is thrown horizontally with an initial speed of 10 m/s from a bridge. If air resistance could be ignored,how long would it take the stone to strike the water80m below the bridge?(A) 1s(B) 2s(C) 4s(D) 6s(E) 8s8. A soccer ball, at rest on the ground, is kicked with an initial velocity of 10 m/s at a launch angle of 30° . Calculate its total flight time,assuming that air resistance is negligible.(A) 0.5 s(B) 1 s(C)1.7s(D) 2s(E) 4s9. A stone is thrown horizontally with an initial speed of 30 m/s from a bridge. Find the stone’s total speed when it enters the water 4 seconds later.(Ignore air resistance.)(A) 30 m/s(B) 40 m/s(C) 50 m/s(D) 60 m/s(E) 70 m/s10.Which one of the following statements is true concerning the motion of an ideal projectile launched at an angle of45°to the horizontal?(A) The acceleration vector points opposite to the velocity vector on the way up and in the same direction as the velocity vector on the way down.(B) The speed at the top of the trajectory is zero.(C) The object’s total speed remains constant during the re flight.(D)The horizontal speed decreases on the way up and increases on the way down.(E)The vertical speed decreases on the way up and increases on the way down.来源于：小马过河小马过河资料下载频道，欢迎您来下载!。

alevelm2知识点总结

alevelm2知识点总结1. KinematicsKinematics is the study of motion and its causes, including the displacement, velocity, and acceleration of objects. In A-Level Mathematics M2, students will learn about concepts such as scalar and vector quantities, displacement, velocity-time graphs, and acceleration-time graphs.Scalar and vector quantities: Scalar quantities have only magnitude, while vector quantities have both magnitude and direction. Examples of scalar quantities include speed and temperature, while examples of vector quantities include displacement and velocity.Displacement: Displacement is the distance and direction of an object’s change in position from its starting point to its ending point. It can be determined using vector notation, which specifies both magnitude and direction.Velocity-time graphs: Velocity-time graphs depict an object’s velocity over a period of time. The gradient of a velocity-time graph represents the acceleration of the object, while the area under the graph shows the total displacement of the object.Acceleration-time graphs: Acceleration-time graphs depict an object’s acceleration over a period of time. The gradient of an acceleration-time graph represents the rate of change of acceleration, while the area under the graph shows the total change in velocity.2. ForcesForces are interactions that cause an object to change its velocity. In A-Level Mathematics M2, students will learn about equilibrium of forces, resolution of forces, and friction. Equilibrium of forces: When the vector sum of all the forces acting on an object is zero, the object is said to be in equilibrium. This means that it is not accelerating and that the forces are balanced.Resolution of forces: Forces can be resolved into components that are perpendicular to each other. By breaking forces down into their horizontal and vertical components, students can analyze their effects more effectively.Friction: Friction is the force that opposes the motion of one surface as it moves over another surface. It can be divided into two types: static friction, which prevents two surfaces from sliding past each other, and kinetic friction, which opposes the relative motion of two surfaces that are sliding past each other.3. MomentsMoments are turning effects produced by forces. In A-Level Mathematics M2, students will learn about the principle of moments, torque, and center of mass.Principle of moments: The principle of moments states that for an object in equilibrium, the sum of the clockwise moments about any point is equal to the sum of the anticlockwise moments about that same point.Torque: Torque is the measure of a force’s tendency to produce rotational motion. It is calculated as the product of the force and the perpendicular distance from the pivot point to the line of action of the force.Center of mass: The center of mass is the point at which the mass of a body may be considered to be concentrated. It is the point at which a single force would act to cause the same motion as the actual forces acting on the body.In conclusion, A-Level Mathematics M2 covers important topics such as kinematics, forces, and moments. Students will gain a deep understanding of these concepts and their applications, allowing them to solve complex problems and analyze real-world scenarios. Mastering these knowledge points is essential for success in A-Level Mathematics and building a strong foundation for further studies in mathematics, physics, and engineering.。

fundamental词根

fundamentalIntroductionFundamental is a word root that comes from the Latin word “fundamentum,” which means “foundation.” It is used to denote something essential, basic, or foundational. In various fields and disciplines, the term fundamental plays a crucial role. Whether it is in science, mathematics, music, or philosophy, understanding the fundamental principles and concepts is vital for building a strong knowledge base.The Fundamentals in ScienceThe Scientific Method: The Fundamental Approach1.Observation: The scientific method begins with observation, wherescientists carefully observe and gather data about a specificphenomenon.2.Question: Based on the observations, scientists develop questionsthat they want to investigate further.3.Hypothesis: A hypothesis is a proposed explanation for theobserved phenomenon. It is a fundamental step in the scientificmethod as it forms the basis for further experimentation.4.Experimentation: Scientists design and conduct experiments totest their hypothesis. These experiments follow specific protocolsand guidelines to ensure accuracy and validity.5.Analysis: After conducting experiments, scientists analyze thedata collected to draw conclusions and determine whether theirhypothesis is supported or refuted.6.Conclusion: The final step involves drawing conclusions based onthe results obtained from the experiments. These conclusions formthe basis for further scientific research and the development oftheories and laws.Fundamental Laws and Principles1.Newton’s Three Laws of Motion:Sir Isaac Newton’s laws ofmotion are considered fundamental laws in classical physics. They describe the relationship between the motion of an object and the forces acting upon it.•Newton’s First Law: An object at rest will stay at rest, and an object in motion will continue moving at a constant velocityunless acted upon by an external force.•Newton’s Second Law: The acceleration of an object is directly proportional to the net force acting upon it and inverselyproportional to its mass.•Newton’s Third Law: For every action, there is an equal and opposite reaction.w of Conservation of Energy: Energy cannot be created ordestroyed; it can only be transferred or transformed from one form to another. This fundamental principle is the basis of manyscientific studies, including thermodynamics and electricity.3.Cell Theory: The cell theory is a fundamental concept in biology.It states that all living organisms are composed of cells, andcells are the basic structural and functional units of life.4.Theory of Evolution:Charles Darwin’s theory of evolution is afundamental principle in biology that explains how species haveevolved over time through natural selection and genetic variation. The Fundamentals in MathematicsFundamental Mathematical Concepts1.Numbers and Operations: Numbers are the fundamental buildingblocks of mathematics. Understanding the properties and operations of different types of numbers (e.g., integers, fractions, andirrational numbers) is essential for mathematical proficiency.2.Algebraic Expressions and Equations: Algebra is a fundamentalbranch of mathematics that deals with the manipulation of symbols and solving equations. It provides a foundation for higher-levelmathematical concepts and problem-solving skills.3.Geometry: Geometry is concerned with the study of shapes, sizes,and spatial relationships. It is a fundamental branch ofmathematics that has practical applications in various fields,such as architecture, engineering, and graphics.4.Calculus: Calculus is a fundamental branch of mathematics thatdeals with rates of change and the accumulation of quantities. It is widely used in physics, engineering, economics, and many other scientific disciplines.5.Probability and Statistics: Probability and statistics arefundamental concepts for understanding uncertainty and makinginformed decisions based on data. They play a crucial role infields such as finance, medicine, and social sciences.Fundamental Theorems and Formulas1.Pythagorean Theorem: The Pythagorean theorem states that in aright-angled triangle, the square of the hypotenuse is equal tothe sum of the squares of the other two sides.2.Fundamental Theorem of Calculus: The fundamental theorem ofcalculus establishes the relationship between differentiation and integration. It states that if a function is continuous on aclosed interval, then the area under its curve can be determinedby evaluating the antiderivative of the function at the endpoints of the interval.3.Fundamental Theorem of Algebra: The fundamental theorem ofalgebra states that every non-constant polynomial equation withcomplex coefficients has at least one complex root.The Fundamentals in MusicMusical Fundamentals1.Pitch and Melody: Pitch is the fundamental characteristic ofsound that determines its highness or lowness. Melody is thearrangement of pitches in a particular sequence, forming a musical line or tune.2.Rhythm and Beat: Rhythm is the pattern of sounds and silences inmusic, while beat is the steady pulse that divides the music into regular units of time.3.Harmony: Harmony refers to the combination of different pitchesplayed simultaneously, creating chords and chord progressions. It adds depth and richness to music.4.Dynamics and Expressiveness: Dynamics refers to the variations inloudness and softness in music, while expressiveness relates tothe interpretation and emotional qualities conveyed throughmusical performance.5.Musical Notation: Musical notation is a system of writing downmusic using symbols and musical symbols, making it possible tonotate and transmit musical ideas over time.Fundamental Musical Practices1.Scales and Modes: Scales and modes are fundamental frameworks fororganizing pitches in music. They provide a structure for melodies and harmonies.2.Chords and Chord Progressions: Chords are formed by combiningthree or more pitches played simultaneously. Chord progressionsprovide a sequence of chords that creates harmonic movement andsupports the melody.3.Form: Musical form refers to the organization and structure of apiece of music. Common forms include binary form, ternary form,and sonata form.4.Improvisation: Improvisation is the spontaneous creation of musicin real-time, often associated with jazz and other improvisational genres.5.Performance and Interpretation: Performance and interpretationinvolve bringing music to life through expressive and skillfulexecution. It requires a deep understanding of the fundamentalaspects of music and artistic expression.ConclusionThe term fundamental is pervasive in various fields and disciplines, from science and mathematics to music and beyond. Understanding the fundamental principles, laws, and concepts forms the building blocks for deeper knowledge and exploration. By grasping the fundamentals, individuals can build a strong foundation to expand their understanding and make meaningful contributions within their chosen fields.。

《物理双语教学课件》Chapter 2 Kinematics 运动学

Chapter 2 KinematicsMechanics, the oldest of the physical science, is the study of the motion of objects. The calculation of the path of a baseball or of a space probe sent from Earth to Mars is among its problem. When we describe motion we are dealing with that part of mechanics called kinematics. When we relate motion to the force associated with it and to the properties of the moving objects, we are dealing with dynamics. We will discuss kinematics in this chapter and dynamics in the following chapters.Usually the motion of an object is very complicated. We will restrict to discuss the motion of a particle, or mass-point.2.1 Position, Displacement, Velocity, and Acceleration1.Position vector r is a vector thatextends from a reference point (usuallythe origin of a coordinate system) tothe object. We can write r as r =x i +y j+z k.Where x i, y j, and z k the vector components of r and the coefficients x, y, and z are its scalar components.The unit of position is meter in SI system.2. Displacement vector r ∆: As an object moves, its position vector changes. If the object has position vector r 1at time t 1 and position vector r 2 at later time t t ∆+1, as shown in above figure. The displacement vector is the difference of the two position vectors r 2 and r 1 during the time intervalt ∆, whichcan be written ask z z j y y i x x k z j y i x k z j y i x r r r )()()()()(12121211122212-+-+-=++-++=-=∆ See sample problem 4-1and checkpoint1 of P54 and P55.3. Average velocity vector : If a particle moves through a displacementr ∆ in a time interval t ∆, then its averagevelocity isk t z j t y i t x t k z z j y y i x x t r v ∆∆+∆∆+∆∆=∆-+-+-=∆∆=)()()(121212 Instantaneous velocity vector : is defined as the limit of the average velocity as the time intervalt ∆becomesinfinitesimally short. We can write it ask v j v i v k dt dz j dt dy i dt dx dt r d t r v z y x t ++=++==∆∆=→∆0lim The direction of the instantaneous velocity of a particle is always tangent to the path of the particle, as shown in the figure.The unit of velocity is 1-⋅s m .The magnitude of the velocity of a particle is its speed . Speedis a scalar quantity.dt r d dt r d v v ===Average speed is the total distance covered by the object over the time interval.t cedis total v ∆=tan4. Acceleration vector : When a partic le’s velocity changes from v 1 to v 2 in a time intervalt ∆, its average acceleration a during t ∆ ist v v t v ∆-=∆∆=12Instantaneous acceleration vector is defined as the limit of the average acceleration as the time intervalt ∆ becomes infinitesimally short. It can be express ask a j a i a k dtdv j dt dv i dt dv dt v d t v a z y x z y x t ++=++==∆∆=→∆0lim The unit of acceleration is2-⋅s m .2.2 One Dimensional Motion with Constant Acceleration.1. The rule of one-dimensional motion with constant acceleration can be derived as follow:200000000)(21)()]([)(sin 00t t a t t v s s dt t t a v ds t t a v v adt dv adtdv therefore a dt dv or c a ce t t v v -+-+=⇒-+=-+=⇒====⎰⎰ We also have:)(2])(21)([2)()(202020002020200202s s a v t t a t t v a v t t a t t a v v v -+=-+-+=-+-+= If we suppose t 0=0, Then we haveasv v at t v s s atv v 2212022000=-++=+= 2. Free-falling bodies: If you throw an object either up or down and could somehow eliminate the effects of air on its flight, you would find that the object accelerates downward at a certain rate. We call it free-fall acceleration g. The object is a free-falling body . The value of g change slightly with latitude and with height. At sea level in the mid-latitudes the value is9.8m/s 2, we will use it to solve the problem in this course. If we choose the upward direction along the y axis, then the free fall acceleration is negative. If we suppose t 0=0 and y 0=0, we will have:202200221v v ay gt t v y gtv v -=-=-=2.3 Projectile MotionIf a particle moves in a vertical plane during free fall, it undergoes projectile motion . The object can be called a projectile . Throughout we will assume that the air has no effect on the motion of the projectile.Projectile motion has the feature that the horizontal motion and the vertical motion are independent of each other. So we can deal with them respectively.If we choose the rectangular coordinate as the following figure, then000000sin cos θθv v and v v y x ==1. The horizontal motion is motion without acceleration. So we havet v t v x x x 0000cos θ==-2. The vertical motion is the motion of free fall. So we have)(2)sin ()sin (21)sin (210200200200200y y g v v gt t v v gt t v gt t v y y y y y --=-=-=-=-θθθ 3. The equation of the path : Let x 0=0 and y 0=0, eliminating the t between the equations above, we get20020)cos (2)(θθv gx x tg y -=It is the equation of a parabola , so the path is parabolic.4. The horizontal Range R is horizontal distance the projectile has covered when it returns to its initial height. We have 021)sin (21cos 200200000=-=-=-=-=gt t v gt t v y y tv x x R y θθ Eliminating the time t, we get02000202sin cos sin 2θθθgv g v R ==2.4 Uniform Circular Motion1. A particle is in uniform circular motion if it travels around a circle or circular arc at constant speed. Although the speed does not vary, it means the magnitude of the velocity is a constant quantity, but the direction of the velocity alters with the time. So the particle is accelerating.2. The feature of uniform circular motion: We have)()()(00t v v t r r t t v s s c v ==-=-⇒= 3. The acceleration of uniform circular motion:(1). Direction: (2). Magnitude: r v a 2=4. Another look at the acceleration of uniform circular motion: 0200000)()(n rv r r v v dt t d v dt t d v t dt dv t v dt d dt v d a =-==+=== 5. The figure shows the relation betweenthe velocity and acceleration vectorsat various stages during uniformcircular motion. Both vectors haveconstant magnitude as the motionprogress, but their directions change continuously. The velocity is always tangent to the circle in the direction of motion; the acceleration is always directed radially inward . Because of this, the acceleration associated with uniform circular motion is called a centripetal acceleration .2.5 Relative motion in two or three dimensions1. The figure shows reference frames A and B, now two dimensions, of course, it can be extended to three dimensions if you would like to. Two observers are watching a movingparticle P in the twodifferent frames. Weassume that the two framesare separating at a constantvelocity v BA (both of themare inertial frames ) and wefurther assume that their axes remain parallel to each other for simplicity.2. Position vectors: If two observers on frame A and B each measure the position of particle P at a certain instant. From the vector triangle in the figure, we have the vector equation BA PB PA r r r +=3. Velocity vectors : If we take the time derivative of above equation, we will get the velocity relation of the particle as measure by the two observersBA PB PA v v v +=4. Acceleration vectors : If we take the time derivative of above equation, we will have a connection between the two measured accelerationPB PA a a =。

physicsI

Chapter 1 KinematicsPhysics deals with matter, motion and their interaction. It is the most fundamental science.Mechanics is a branch of physics dealing with mechanical motion of subjects. In kinematics, we investigate the description of motion.Motion is relative. Reference frames are always needed to specify the motion. To make quantitative description, coordinate systems are necessary. Rectangular coordinate system consists of three mutually perpendicular X, Y and Z axes which intersect at the origin O .To describe the motion of a particle, we draw a vector r extending from the origin of the coordinate system to the particle’s position. It is called position vector. Thusk j i r z y x ++= (1)in which i, j and k are unit vectors and x, y and z are the components. The components may be positive, negative and zero.Mechanical motion is defied as the process of change in position with time. The position vector can be correlated with the time by means of vector function)(t r r = (2)Its three components are written by the following scalar functions)();();(t z z t y y t x x === (3)The path equation can be obtained by eliminating t from Eqs.(3). If the path of the particle is a straight line, the motion is called as a rectilinear motion; if the path is a curve, the motion is called as a curvilinear motion.The displacement is the change in position during a given time interval. SoP Q r r r -=∆ (4) Displacement is a vector, its magnitude r ∆ is the chord PQ; path is a scalar ⋂=∆PQ s . In most case,s ∆≠∆r . You should be aware of the difference between r ∆ and r ∆.If r ∆ is the displacement that occurs during the time interval t ∆, its average velocity for this interval is defined ast ∆∆=r v (5)To describe the motion of a particle at a given time t or at a given point P , we must make t ∆ very small. The instantaneous velocity at any time t is obtained by evaluating t ∆∆/r in the limit that t ∆approaches zerodt d t t r r v =∆∆=→∆0lim (6)Therefore, the instantaneous velocity is defined as the time derivation of the position vector. Velocity is a vector tangent to the path, and points to the advance direction. The three components of velocity is given by;;;dt dz v dt dy v dt dx v z y x === (7)Similarly, the path length s ∆ divided by the time taken t ∆ is called the average speed, so dt ds /is the instantaneous speed. Note that the speed is a scalar. The magnitude of instantaneous velocity equals instantaneous speed.The change in velocity during the time interval t ∆ is represented by P Q v v v -=∆ in the vector triangle. The average acceleration is defined byt ∆∆=v a (8)Using the same method as in definition of velocity, the instantaneous acceleration at time t is defined by dt d t it t v v a =∆∆=→∆0lim (9)which is the time derivation of velocity vector. From Eq.( 6), we can also write Eq.(9) in the form22dtd dt d r v a == (10) So the three components of acceleration are given by;;;222222dtz d dt dv a dt y d dt dv a dt x d dt dv a z x y y x x ====== (11) and the magnitude of the acceleration is222zy x a a a a ++= (12) Acceleration vector has the same direction as the limit direction of change of velocity when 0→∆t , which is always pointing toward the concavity side of the path. We can decompose acceleration into two perpendicular components: tangential acceleration t a and normal acceleration n a . The change in the magnitude of velocity is related to the tangential acceleration and change in direction of velocity is related to the normal acceleration. We haveρ2;v a dt dv a n t == (13) where ρ is the radius of curvature.。

1、下载文档前请自行甄别文档内容的完整性，平台不提供额外的编辑、内容补充、找答案等附加服务。
2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
3、如文档侵犯您的权益，请联系客服反馈,我们会尽快为您处理(人工客服工作时间：9:00-18:30)。

1. DOF of a Mechanism
Example:
1. DOF of a Mechanism
1.4 Points of Attention to Calculating DOF A:Multiple joints(compound hinges) B:Passive DOF C:Redundant Constraints
2.1 Basic Concepts of Mechanisms
Degree of freedom (自由度自由度) 自由度 Link （构件）构件） Kinematic Pairs (Joints)（运动副）（运动副） Machine Element（机械零件） Element（机械零件） Fixed link ( frame)（机架）（机架） Crank, couple,rocker（曲柄、连杆、摇杆）（曲柄、连杆、摇杆） Kinematic Chain（运动链）（运动链） Mechanism（机构）（机构） Machine（机器）（机器）
Diagram of Kinematic pairs（运动副简图）（运动副简图）
Prismatic joint Revolute joint Cylindric joint Spherical joint Helical joint
Diagram
Type DOF
P joint 1
R joint 1
2.1Basic Concepts of Mechanisms(4)
Lower pairs（低副）:joints with surface contact （低副）
Prismatic joint Revolute joint Cylindric joint Spherical joint Helical joint
C joint 2
S joint 3
H joint 1
2.1Basic Concepts of Mechanisms(8)
Machine Element（机械零件）: minimum （机械零件） unit during manufacturing
This link has several machine elements Binary link
DOF
5
2
2.1Basic Concepts of Mechanisms(6)
Examples of higher pairs
Form –closed Force –closed need external force closed by its geometry
2.1Basic Concepts of Mechanisms(7)
nodes nodes
Binary link
Quaternary link
2.1Basic Concepts of Mechanisms(3)
Joint (铰链或运动副)
Joint: is a connection between two or more links(at their nodes), which allows some motion, or potential motion, between the connected links. Joints (also called kinematic pairs) can be classified in several ways: 1. By the type of contact between the elements, line, point, or surface. 2. By the number of degree of freedom allowed at the joint. 3. Others
Chap.2 Kinematics fundamental
2.1 Basic Concept 2.2 Determining Degree of Freedom 2.3 How does a kinematic chains transform into a mechanism? 2.4 Kinematic Diagram of Mechanism 2.5 Principle of Mechanism Constitution
2.1Basic Concepts of Mechanisms(1)
Degree of Freedom（自由度）（自由度）
The system’s DOF is equal to the number of independent parameters which are needed to uniquely define its position in space at any instant of time.
2.2 Determining Degree of Freedom
1. Degree of Freedom of a Mechanism 2. Kinematic Diagram of Mechanism
1. DOF of a Mechanism
1.1 Review the definition of DOF 1.2 Degree of freedom of a link 1.3 DOF formula of a planar mechanism 1.4 Points of Attention to Calculating DOF
1. DOF of a Mechanism
1.1 Definition DOF : The number of independent coordinates required to define its position. 1.2 Degree of freedom of a link

Diagram of Links and Joints
C: Redundant Constraints
2.3 How does a kinematic chains
transform into a mechanism?
Conclusion: 1) At least one link has been grounded 2) DOF equals to the number of the drivers
2.1Basic Concepts of Mechanisms(9)
*Kinematic Chain（运动链）: an assemblage of links （运动链） and joints. If these links are connected in such a way that no motion is possible, a locked chain(structure) result. Mechanism（机构）: a kinematic chain in which at （机构） lease one link has been “grounded”, or attached, to the frame of reference(which itself may be in motion); the links in the kinematic chain move upon each other with definite relative motion. Machine（机器）: a combination of resistant bodies （机器） arranged to compel the mechanical forces of nature to do work accompanied by determinate motion.
Characteristics of a machine:
1. Composed of man-made elements 2. The elements are arranged to transmit motion 3. Can transmit energy, and fulfill certain work for human being. Constitution:
Planar joint
Type DOF
P joint 1
R joint 1
C joint 2
S joint 3
H joint 1
F joint
3
2.1Basic Concepts of Mechanisms(5)
Higher pairs（高副）:joints with point or line contact （高副）
2.1Basic Concepts of Mechanisms(11)
Coupler 连杆 Crank 曲柄
Fully rotating revolute; Partially rotating revolute;
Frame or Ground 机架
Rocker 摇杆
Planar mechanism: all links of a mechanism move in planes that remains parallel to each other. Otherwise, it is a Spatial mechanism.
2.1Basic Concepts of Mechanisms(2)
Link（构件）:is a rigid body which possesses （构件） at lease two nodes which are points for attachment to other links.
Ternary link
Crank(曲柄 a link which makes a complete revolution 曲柄): 曲柄 and is pivoted to ground. Rocker（摇杆）: a link which has oscillatory(back and （摇杆） forth) rotation and is pivoted to ground. Coupler(connecting rod，连杆): a link which has complex motion and is not pivoted to ground. Ground（机架）: any link or links that are fixed （机架） (nonmoving) with respect to the reference frame. ( Note that reference frame may in fact itself be in motion.)