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Thread: moment diagram with weight transfer

  1. #141
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    Weight a Minute

    For weight analysis, sure do the summation of point masses in a spreadsheet. Inertia, rig the tri-filer pendulum. Engine and transmission assembly is not a point mass and it alone could account for 1/2 the total inertia of the whole car. Plus, it's inertia ellipsoid's axes probably won't line up nicely with the car coordinates you choose. There are some other ways to get these values, including methods for obtaining unsprung inertias.

  2. #142
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    Ralph,

    The big problem, for me, has always been coming up with K short of building a very large trifilar pendulum.
    Is there a way to reasonably estimate the second moment of mass about the Z axis?
    1. The rough calculation by summing contributions from component masses (ie. ~Sum-all-i's(MoI.i + Mi x Ri^2) is the quickest, easiest way, albeit probably at +/-10% error (?). But 90% of an answer is better than no answer. Doing this is also good from a "design" viewpoint, in that doing the calcs many times over makes it clear how big the influence is from masses at the extremities of the car (because of R-squared).

    2. Here is my first post on an earlier thread discussing other methods. There is a simple sketch on the next page (p3) showing the "Compound Pendulum" method of finding the car's Yaw-Inertia (the appropriate equation is a few posts before that sketch). But this method is probably better suited to smaller FS/FSAE cars, and maybe not your full-sized ones? Although years ago I did measure some off-road-buggys' CG-Heights and Yaw-Inertias by "swinging" them from a tree branch! (As noted, windy days = !!! !!! )

    3. Another method is becoming commonplace nowadays, probably because even easier. You build a platform a bit like a playground merry-go-round, or a "turntable". The car sits on the turntable with car-CG directly above the vertical axis. The turntable has a spring between itself and the floor that provides the rotational "restoring force" (measured as torque/rotation-angle, say N.m/radian). You then set this rotational spring-mass system oscillating and do the fairly simple maths to get Yaw-Inertia of the car alone (ie. subtract the turntable's MoI).

    I reckon said turntable could be as simple as a front-wheel+axle-bearings+upright assembly. The (steel) wheel sits on its side on the floor, with the upright uppermost. Bolt a very simple frame (angle-iron or welded-RHS...) to the upright. Gently lower the car onto this frame, with car-CG directly above axle (an offset CG will tilt the whole assembly, so adjust car position until it sits level). The "restoring force" spring might just be an appropriately sized steel coil-spring working in both tension and compression, and mounted between car-nose and heavy-object-sitting-on-floor. You give the turntable+car a small rotation from its zero-position, let go, and measure time for ten rotational oscillations. Divide by 10 for period "T", and do the maths...
    ~~~o0o~~~

    Wil,

    I'd been learning (or more accurately, trying to learn) about the effects of rotating a rigid body about a non-principal inertial axis, ... (the inertia tensor has ... non-zero products of inertia).
    Yes, despite the extreme simplicity of Newton's Laws (ie. little more than F -> P-dot), once you glue a few of N's particles together into a single "rigid body", and then get that body rotating, it all get very interesting, very quickly!

    Here is a short thread where I discuss Principal-Axes... about half-way down the first page, and clarify again at top of 2nd page. (Wow, ten years ago...). Note that many different issues are discussed on that thread, and not always very clearly because of the difficulty of trying to explain in words what really needs a lot of pictures.

    The gist of what I was getting at in above thread, is that even in the very simple "Steady-State" cornering shown in Sketch-3 above, WITH the "tangent condition" (ie. the "centrifugal force" at CG is exactly perpendicular to car-centreline, so it is purely "lateral" to the car), it is possible for the car to have a Dynamic LONGITUDINAL-load-transfer (ie. F:R "axle-weights" are different to when car stationary on scales)! This happens whenever the car's principal-axes are not "square" with its rotation W-vector (eg. longitudinal principal-axis NOT horizontal).

    The dumb-bell analogy I gave is not completely accurate, but it gives a reasonably clear explanation of the effects. For a completely accurate way of turning a 3-D Eulerian problem into a Newtonian problem of "point-masses connected by massless-rods", you only need a slightly more complicated "dumb-bell".

    Z
    Last edited by Z; 10-09-2015 at 11:18 PM.

  3. #143
    Ralph,
    1. The rough calculation by summing contributions from component masses (ie. ~Sum-all-i's(MoI.i + Mi x Ri^2) is the quickest, easiest way, albeit probably at +/-10% error (?). But 90% of an answer is better than no answer. Doing this is also good from a "design" viewpoint, in that doing the calcs many times over makes it clear how big the influence is from masses at the extremities of the car (because of R-squared)
    .

    Z,

    I will take a 10% error over having nothing at all.

    After all, in almost all cases I am interested in the relative size of some effect, not calculating an exact number for the purposes of chassis tuning.


    The dumb-bell analogy I gave is not completely accurate, but it gives a reasonably clear explanation of the effects. For a completely accurate way of turning a 3-D Eulerian problem into a Newtonian problem of "point-masses connected by massless-rods", you only need a slightly more complicated "dumb-bell".
    Hmm...a more sophisticated dumb-bell...thinking. The first thought that comes to mind is that the dumb-bell must be viewed in both side and plan with the angle of the principle axis in side view being accounted for in the plan view. But that is probably me just being simple minded again.

    Off topic but...

    Check this link to a Dover book on 'The Variational Principles of Mechanics' by Cornelius Lanczos. Click the Google preview button and go to his section on de'Alembert. I would be interested in your thoughts.

    http://store.doverpublications.com/0486650677.html

    Also one of my old texts 'Analysis and Design of Mechanisms' by Deane Lent circa 1970 covers the construction of your General Planer Acceleration Diagram quite well for draftsman working out linkage accelerations on a drawing board. Very basic book, think it lists at about five U.S. dollars these days, but I use it quite often as a reference in solving graphic vector problems.

    Lastly, a question.

    When the kids on the bus applied an input to the driver to cause him to initiate his 'step steer' input how did you estimate:

    1. The amount of steer angle input?
    2. The associated V to Tire centerline angle (i.e. 'slip angle')?
    3. The forces generated at the front tires for your chosen steer angle? (Remember a post way back where you mentioned to someone generating representative curves for tire data)


    In closing, the sound of the crickets on this thread is deafening.

    Ralph
    Last edited by rwstevens59; 10-10-2015 at 03:41 PM.

  4. #144
    Z,

    It is probably not of much benefit to spell out the above geometric method in detail here. The same results can be obtained with the more conventional Euler's Equations methods that are usually taught these days, and which are very briefly outlined in next post.

    However, it is well worth closely studying the overall pattern of A-vectors here, and the fact that some features of this pattern are ALWAYS PRESENT. These general features covered next.

    I, for one, would find it informative I'm sure. But, as my 'Z binder' is turning into a textbook size, I realize one can only ask for a finite amount of 'free' education.

    Ralph

  5. #145
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    Ralph,

    Hmm...a more sophisticated dumb-bell...thinking.
    One helpful way of understanding 3-D Dynamic problems is to reduce the complicated 3-D body down to three "rigidly connected" dumb-bells that intersect at 90 degrees to each other at their mid-points (which are at the body's CG), and with each point-mass at the ends of the dumb-bells = Mt/6 (because six point-masses). Each dumb-bell is aligned with a Principal-Axis, and their three lengths are adjusted to match the three respective MoIs of the "real body" about its Principal-Axes. (Note that "MoI.X" = MoI-of-BOTH-Y-AND-Z-dumb-bells about the X-axis = (Mt/3) x (Ry^2 + Rz^2), where Ry is half-length of Y-dumb-bell, etc.).

    It is quite fascinating that the dynamic behaviour of ANY body, no matter how complicated its mass-distribution, can ALWAYS be modelled with such a simple "three dumb-bell" shape (assuming reasonable rigidity). For example, consider that the most complicated piece of "modern art" (= pile of steel junk welded together and dumped in town-square) ALWAYS has three mutually perpendicular axes within it, about which it can be spun with perfect dynamic balance (ie. no wobbling). Amazing!

    (Edit: It is amazing that there is a SINGLE axis about which any odd-shaped body can be spun perfectly smoothly. More amazing that there are always three such axes (at least), and they are always exactly perpendicular to each other!)

    In Sketch-2 I could also have found the AP by considering the bus's mass-distribution to be a simpler, symmetric, dumb-bell with equal masses Mt/2 at each end of a massless-rod 2 x K long. I would then have had to decompose the total force Ft into two separate forces, F1 and F2, both parallel to the original Ft, and acting at the two point-masses. Then work out each mass's acceleration (ie. proportional to its force), then draw a line through the tips of the two A-vectors to find the position of zero acceleration, namely the AP.

    In the case of the Sports Bus, the Ft force lies "outside", and in front of, the dumb-bell, giving a rightward F1 on the forward dumb-bell half-mass, and leftward F2 on the rearward half-mass. So the AP lies "inside" the symmetrical dumb-bell, or "K-ring". In the case of the Slow Coach the Ft force lies between the two half-masses, so both forces and accelerations are rightward (but greater on forward half-mass), so the AP lies outside and behind the dumb-bell or K-ring.

    More on this below...
    ~o0o~

    Check this link to a Dover book on 'The Variational Principles of Mechanics' by Cornelius Lanczos. Click the Google preview button and go to his section on de'Alembert. I would be interested in your thoughts.
    Well, I was disappointed that he starts that section with "... the fundamental Newtonian Law of Motion ... mA = F"!!!

    I will keep stressing that the gist of N's LoMs is "... an impressed force causes a change in the quantity of motion...", with Newton defining "quantity of motion" as "the velocity and quantity of matter conjointly." (ie. what we call "Momentum" nowadays). So "F causes P-dot". Thinking in terms of momentum is much more helpful in solving all these problems, especially when things start rotating!

    Otherwise that section of the book seems to confirm that d'Alembert's Principle works just fine, everywhere. Without exception!
    ~o0o~

    Lastly, a question.

    When the kids on the bus applied an input to the driver to cause him to initiate his 'step steer' input how did you estimate:

    1. The amount of steer angle input?
    2. The associated V to Tire centerline angle (i.e. 'slip angle')?
    3. The forces generated at the front tires for your chosen steer angle?
    The overriding concern when doing these sketches is not so much "solving the given problem", but more a question of how to fit-in all the necessary information on one sheet of paper, preferably without overcrowding in one part of the sketch and large blank spaces elsewhere.

    So at the very first "rough draft" stages I was trying to find a reasonably realistic sized bus, travelling at a realistic speed, and cornering around a realistic corner radius, preferably with the bus AND corner centre visible on the same page (as in Sketch-3 and 4). And I also wanted the various vectors to be at a scale where the smaller ones are visible, but the larger ones are not shooting off the edge of the page. And where something important like the AP is not in an extremely crowded part of the sketch...

    So, in Sketch-2 the front-steer/slip-angles and the amount of positive-Ackermann, etc., was chosen simply to put the "Flf + Frf = Ft" vector parallelogram in an un-crowded part of the page! The steer/slip-angles of ~15 degrees are on the high side, but they make clear that the front-wheel forces are hardly ever purely sideways to the bus, and this then puts M1 and M2 (= the AP) off the bus's centreline, and so on.

    The perhaps unrealistically large tyre-forces I used, based on tyre-road Mu = ~1.5, were chosen so I could get a large enough Yaw-Rate W to give visibly large enough Centripetal accelerations in Sketch-4. As it turned out, the A-vector of mass M1 is just long enough to be visible, at about 3 mm long on the A4 sheet of paper (it appears as a short arrowhead "Am1" at the bottom of M1).

    But most importantly here, the position of the AP in Sketch-2 is independent of the magnitude of the force Ft acting on the bus. It depends only on the relative positions of the LoA of Ft and the CG, and on the Yaw Radius of Gyration K. I give the standard derivation of this below.
    ~o0o~

    ...to spell out the above geometric method [used in Sketch-4] in detail here.
    Very briefly, the point-mass M2 is put on the LoA of the total EXTERNAL force Ft, with the bus's CG perpendicular to the LoA at M2. But M2 also has an INTERNAL force acting on it, coming from M1 via the "massless rod", with this force due to the bus's already pre-existing Yaw-Velocity W. So M2's A-vector "Am2" is NOT parallel to Ftotal, but has a small centripetal component towards CG (ie. toward top of sketch). The size of this centripetal component is worked out using the known W (= 1 rad/sec) and distance between M2 and CG (ie. A.centripetal = W^2 x R). Or it can be worked out using known W, and distance between M1 and M2, and relative masses of M1 and M2.

    KEY STEP! -> In bottom-right of Sketch-5, the angle the "At+An" vector makes with the R-vector (ie. the angle shown as "((" ), is the same as the angle ALL A-vectors make with radii drawn from the AP. In vector terms "At+An" = A2 - A1. So, back in Sketch-4, add MINUS-Am1 to Am2, which moves the tip of Am2 up about 3 mm... Now the angle between this new vector "Am2 - Am1" and the line joining M2 and M1 is the angle between ALL A-vectors and radii from the AP.

    I did the above geometric construction of the Am1 and Am2 vectors at a larger scale, on the back of a bit of cornflakes-packet cardboard. The angle came out to about 76.6 degrees (my drawing-board protractor gives that level of accuracy). I used scissors to cut out a wedge of the cardboard with that included angle. Then simply a matter of placing one edge of this wedge along Am2 and drawing a pencil-line at ~77 degrees to Am2. This pencil line must pass through the AP. Then another line at 77 degrees to Am2 (Am2 necessarily has direction M1-M2)...

    BINGO! The AP is at the intersection of these two lines. Then use the cardboard wedge to draw the directions of all the other A-vectors, all at 77 degrees to radii from the AP. Then measure out magnitudes of all A-vectors by direct proportion to, say, the scale of the Am2 vector (which is also easily done with a triangle of cardboard). DONE!

    (Note: Just a little extra thinking required to make sure that all the "77 degrees" are going the right way from the radii, but this is fairly easy...)
    ~o0o~

    (Bit more coming... 10k char limit!)

    Z
    Last edited by Z; 10-14-2015 at 08:58 PM. Reason: More details...

  6. #146
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    CENTRE OF PERCUSSION & CENTRE OF OSCILLATION - <- Google these terms and you find essentially the same problem as in Sketch-2.

    For the School Bus the Centre of Percussion is the point M2, namely the point on the LoA of the force Ft acting on the (non-rotating) body, with the body's CG being perpendicular to the LoA at this point. We take it as given here that the distance from CG to Ft's LoA (or M2) is "A".

    We seek the Centre of Oscillation (this term coming from pendulum theory, and = AP here), namely the point about which the whole Body (= bus) starts to rotate when first acted on by the force. We take it as given that the Body has total mass = Mt, Radius of Gyration = K, and lives in Flatland.

    Step 1. We first convert the single linear force Ft into an "equivalent force system" of another force Ft', of same magnitude and direction as Ft but acting on a LoA through the CG, together with a couple T = Ft' x A acting anywhere on the Body.

    Step 2. By Euler's Rigid Body Equations (student-special "Flatland-lite" version), force Ft' causes rate-of-change of Linear Momentum (Mt x V) of the Body, and couple T causes rate-of-change of Rotational Momentum (Mt x K^2 x W) of Body. Or, assuming Mt and K are unchanging, and taking "=>" to mean "causes",
    Ft' => Mt x V-dot, and
    T (= Ft' x A) => Mt x K^2 x W-dot.
    In words:
    2a. Force Ft' Linearly Accelerates every point of the whole bus, in the direction of Ft', at rate A = V-dot = Ft'/Mt (like top-left of Sketch-5).
    2b. Couple T Rotationally Accelerates every point of the whole bus, in Clockwise direction about CG, at rate W-dot = (Ft' x A)/(Mt x K^2) (like top-second-from-left of Sketch-5).

    Step 3. The acceleration of any point of the bus (wrt absolute space...) is simply the vector addition of the above two patterns. Since, in rotating motion V-dot = W-dot x R (where R is the radius vector, from-centre to-given-point, and this much dumbed down here...), the pattern of 2b above gives at any point a linear acceleration of V-dot = R x (Ft' x A)/(Mt x K^2).

    We seek the point where the acceleration is zero (ie. the AP), which is the point where the two vectors in 2a and 2b are equal and opposite. So, we seek Ro such that,
    Ft'/Mt = -Ro x (Ft' x A)/(Mt x K^2).

    The force Ft' and the total mass Mt cancel out (so easy!), leaving K^2 = -Ro x A. And writing -Ro as the more usual "B" gives,
    K^2 = A x B... ALWAYS!

    Or..., the position "B" of the Acceleration Pole is dependant ONLY on the distance "A" from CG to the acting force (= ~distance from CG to front-axle of the bus), and on the Radius of Gyration "K" of the bus.

    Or..., the position of the AP is NOT influenced by how hard the bus driver pulls on the steering-wheel, because the bus ALWAYS starts rotating about the same point near the rear-axle, regardless of the size of the front-wheel forces. (Err..., not considering changing live loads, or boisterous school kids!).

    Z
    Last edited by Z; 12-02-2015 at 07:08 PM. Reason: Clarifying many details... & fix splenig!

  7. #147
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    I received the following PM recently. I reply here so that other students with similar questions can see it.

    Re: Turn Radius and Yaw Velocity.

    ... School Bus Example, I have a question, if you could help me:

    In order to calculate the Turn Radius and the Yaw Velocity: should I use the kinematic approach, conecting the perpendicular line of the slip angles and measuring this distance(turn radius), or should I use a dynamic approach, using LatAcceleration = LatForce/mass, and Turn Radius = Velocity/Lateral Acceleration?

    Because I think I can find a turn radius in both cases, but which one should I use when? I make this question for MMM reasons(modeling).

    Thank you very much, it will help me a lot!
    This problem is indeed a deep one. Because, sadly, the problem boils down to a complete lack of foundational education. The only way I can think to (start to) resolve this problem is to restress some of the things I wrote earlier. So...
    ~o0o~

    Classical Mechanics (CM) is a sub-field of APPLIED MATHEMATICS!

    It is NOT a part of Engineering, or of Physics, or of any other "Sciences". However, it can be used to considerable benefit by Engineers, Scientists, and so on.

    CM deals either with idealised "point masses" (Newton), or idealised "continuum mass distributions" (Euler). A mountain of empirical evidence suggests that "reality" is NOTHING like "point masses" or "continuums". In fact, the overwhelming evidence is that at its core "reality" is downright weird. The word "spooky" is most often used to describe this level of reality (see any textbook on Quantum Mechanics).

    Nevertheless, for any system bigger than a small bunch of "atoms" (whatever they are, particles or waves?), Classical Mechanics gives results that are extraordinarily accurate, far more accurate than most any Engineer or Scientist can ever measure. This accuracy, plus CM's core simplicity, is what makes the "CM-model" so useful to everyday usage by Engineers and Scientists.

    But even more important than this practical utility is that CM is a Mathematically rigorous "Axiomatic-Deductive" approach to thinking about the given subject (namely, how does reality work?). (Note here that "A-D" is the modern name for this approach, but, for historic reasons, it is perhaps better described as "Geometric". This "Geometric" approach is about a lot more than just triangles! Google Plato or the Pythagoreans, for a start.)

    Perhaps even more important than the "Axioms" and "Deductions" just mentioned, is that this whole "clear thinking" approach STARTS with very clear DEFINITIONS of all the relevant terms. Only after these Definitions are given, follow the very clear statements of what will be ASSUMED (ie. the "Axioms", or Euclid's "Postulates and Common-Notions"). And then follow, in a clear, incontrovertible, step-by-step way, the Deductions of lots of useful factoids (which, to repeat, must be derived ONLY from the Definitions and Axioms!).

    The gist of what I am getting at here, is that Classical Mechanics, when it is taught properly, is always, and entirely, CRYSTAL CLEAR!

    The whole edifice of CM is complete and consistent. Every question has its answer. There are NO "unknowables". Even the relatively modern development (~1900, Poincare) of Chaotic Dynamics gives exactly predictable "clockwork" results. (The apparent "unpredictability" of chaotic-systems, such as the weather, is merely a by-product of the imprecision of real measurements.)

    The really disappointing thing is that this crystal clear "Geometric" approach to thinking has long been abandoned.

    Nowadays students are just fed a random selection of the factoids (= "Deductions" above), usually in the form of poorly explained equations (= alphabet-soup), which they are expected to believe and understand even though they are given no supporting or foundational explanations of how these factoids work, or why they must be true.

    And, even worse, there then follows a gradual process whereby the miscellaneous factoids are corrupted and perverted (ie. errors creep in), until the whole subject is reduced to gibberish (eg. you are told to "mind your migrating RCs", "use the parallel-axis theorem", etc.!).
    ~~~o0o~~~

    Anyway, enough ranting, and some brief clarifications for above student.

    "In order to calculate the Turn Radius and the Yaw Velocity: should I ..."

    1. "... use the kinematic approach, conecting the perpendicular line of the slip angles and measuring this distance(turn radius),"

    This only works if you ASSUME the car is a rigid body moving wrt a ground-frame (a reasonable assumption). If so, then the "turn-centre" (or IC, or Velocity-Pole) is found directly from its DEFINITION (= ~place with zero relative velocity). And then, of course, the "turn radius" depends entirely on which point of the body you are considering!

    2. "... or should I use a dynamic approach, using LatAcceleration = LatForce/mass,"

    This equation has NO direct connection with a car going around a corner! It could be a body (or point-mass?) accelerating laterally, WITH NO ROTATION AT ALL.

    3. "... and Turn Radius = Velocity/Lateral Acceleration?"

    Well, Velocity-squared! But this equation is based on the ASSUMPTION of a point moving with tangential-Velocity along a path with a given Radius-of-curvature. Why bother "calculating" the Radius ... when it is given to you?

    Also, you will never find a Yaw Velocity from this equation because it applies to a point, and a point has no "front", no "rear", no means to establish "direction", so it cannot "rotate". (See Euclid, Book 1, Page 1, Definition 1, "A point is that which has no part.".)

    Lastly,

    "... I make this question for MMM reasons(modeling)."

    I have never done any MMM (or YMD) modelling, and doubt I ever will.

    But I understand Claude teaches this method. I note he has been very silent on this subject recently, on this and other related threads.

    So, Claude, in return for the student's money, do you not feel an obligation to teach this subject clearly? And, preferably, in accordance with CM?

    Z
    Last edited by Z; 12-04-2015 at 11:01 PM.

  8. #148
    Z,

    Got an even easier way to do this with the schoolyard turntable. Drive the car onto the turntable, attach a spring scale to the unused holes in the license plate, measure the static friction of the assembly (moment required to start turning it) and the dynamic friction of the assembly (moment required to keep turning it at a constant rate) march around with as close as you can to perpendicular with a constant force on the scale, measure angular acceleration of the turntable. Then drive the car off the turntable and put the same moment on the empty turntable and measure its angular acceleration. For an even more accurate measurement stack weights at the center of the turntable to represent the weight on the bearings with a much lower Izz...

    Given that most cars won't break when you hop an FIA curb I think a car's hub and wheel bearings would be adequate here.

    -CPK
    Last edited by Charles Kaneb; 12-06-2015 at 09:20 PM. Reason: Friction!!!
    Charles Kaneb
    Magna International
    FSAE Lincoln Design Judge - Frame/Body/Link judging area. Not a professional vehicle dynamicist.

  9. #149
    Quote Originally Posted by Z View Post
    I have never done any MMM (or YMD) modelling, and doubt I ever will.
    Maybe you should tried it, its classical mechanics.

    All the questions you discuss have to do with the tire force and calculations of slip quantities.

  10. #150
    Well I might be the wrong person to do this since I am measuring my cumulative exposure to vehicle dynamics in months rather than years but, nonetheless, I'll take the bait and answer the questions posed to the best of my understanding.

    --------------------------------------------------------------

    "1. Is LART (= Lateral Acceleration Response Time), all by itself, really the most important metric for a transiently sporty car?

    2. What about YART (= Yaw Acceleration Response Time)!?

    3. Does a forward-biased weight-distribution really maketh a sportier car (as claimed by others, earlier), and if so, then WHY!?

    4. Is the generally faster LART of high-F% cars merely by-product of the "LA" being measured closer to the front-wheels, which is where, on standard "front-steered" cars, this whole transient cornering process is initiated?

    5. Will a high-R% car, such as a Porker, EVER win a race, and why?

    6. What general correlations can be expected between changes to the F:R% of a car, and the consequent changes to its LART and YART, if "all other things are kept equal"?

    7. How big an influence to transient cornering is the car's "K", namely its "Yaw-Radius-of-Gyration" (or "Yaw-MoI")?

    8. Considering the Sports Bus and Slow Coach of Sketch-2, what can be expected of their respective time-histories of F&R tyre-forces as they move from Sketch-2 to Sketch-3, and are these bus's names thus appropriate?"


    ---------------------------------------------------------------------------

    1) No. Clearly a helpful measure but it is dependent on other factors.

    2) At least as important. A fast YART will lead to faster development of slip angles in the rear and a quicker LART.

    3) Not necessarily. An obvious factor in this that was brought up was location of measurement for LART is critical. An important correlation is that would be that if the COG is more forward the reading at the COG will read higher. Perhaps a more universal way to compare between vehicles (or different COG locations)would be to measure LART (directly or with transforms) at the centrepoint of the wheelbase. That being said for analysis of a single car knowing the LART at the COG is likely more helpful.

    4) As in 3.

    5) Obviously they can. One distinct advantage in cornering is the potential for higher instantaneous angular acceleration about the COG. As the COG moves backwards the moment arm from the front tires to the COG becomes longer. With tires that have a constant coefficient of friction this results in a 50/50 distribution being optimal. With real tires, that have a decreasing coefficient of friction with higher normal force, the point of maximum moment and therefore maximum anglular acceleration in the event of a step steer moves backwards. The attached graph shows this for a 1N car with a coefficient of friction of 1 at 0 load.

    Rear weight percentage has a couple other significant benefits outside of cornering. Overall braking should be improved as it more evenly distributes normal forces to the tires, which with all else equal should improve overall traction. In cars with enough torque to cause wheel spin acceleration will also be significantly improved due to higher overall traction force on the drive tires.

    6) As in 5. I would expect higher YART with a slight increase in rear weight percentage (optimized for whatever the normal-force to coefficient of friction response your tires have). More angular acceleration in step-steer should cause quicker slip-angle creation in the rear of the tires and higher LART.

    7) Obviously huge. Lower K means higher angular accelerations, faster YART, and therefore faster LART. Faster response overall means faster generation of cornering forces.

    8) I am not sure I am interpreting this question the correct way. Assuming the bus’s mass is the same in sport mode or slow coach mode, the actual forces at any given geometric point in a turn will be the same. Looking at the two modes over the same timeframe we would see the slow coach develop slip angle in the rear much slower, and from steady state to corner exit the reverse would be true. The sport bus should “snap” out of the turn much quick due to the lower energy required to change the angular momentum and the faster develop of slip angle change.

    --------------------------------------------------------

    Other comments:
    Much of what was shown is obvious from rigid body mechanics, and likely very apparent to most people reading this, but I at least really appreciated the concise explanation of a geometric approach. It is also helpful in thinking about the effects of F/R weight distribution

    From my perspective it seems like the geometric approach Z showed has more commonalities than differences from the MMM. Both allow analysis of balance at varying time and a way to visualize the handling of a car at any given moment in a turn. For any students who have not read the second SAE paper on MMM (SAE 800847) I would highly recommend it. I found it gave a bit more context than RCVD on the development of MMM that was helpful with applying the technique.


    A slight rear weight bias might be something to seriously consider for FSAE. We are running a hybrid configuration that has some other complicating requirements but I justified a rear weight bias based on a couple factors:
    1) Most critically was to maximize rear traction in braking. In conjunction with larger tires a rear weight bias will allow us to maximise regen. Obviously not a factor for many teams. Might be something to consider for teams running really small brakes. It is much easier to fit big differential brakes than big front brakes.
    2) Maximize acceleration. The hybrid powertrain provides significantly more torque than the tires can handle. A better option may have been to downsize the engine for lower weight (already a 250cc). This will likely be very similar to any high power FSAE team.
    3) Encourage neutral handling with larger rear tires. For the reasons above 8” tires are used on the rear with smaller tires up front. More rear weight means less understeer from the higher rear grip.
    4) The final argument (which I will not take credit for using in my design) would be the one discussed above with regards to improving YART by a slight shift of the COG backwards. While the graph I provided demonstrates this effect the actual results will obviously vary significantly with varying tire properties. From a first look it would appear the advantage is likely to only be realized with about a 3-5% shift rearward (Unless your tires have a huge drop in friction with load).

    ---------------------------------------------
    A couple notes on the plot. It was assume that K remained the same regardless as to CG position. With respect to Z's pictures this would be the same as keeping the dumbell rod the same length but shifting it rearward.

    The formula for moment about COG was:

    M = (1-F%) * Wheelbase * (F% * Weight)*Cf

    Where Cf=1 for the baseline and Cf=(1 - .05*Front Weight) for the "real" tire model.
    FR Distribution.jpg


    -Geoff
    University of Victoria Formula Hybrid
    Suspension and Chassis Lead 2014-2016

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