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Thread: Jacking force

  1. #81
    Senior Member
    Join Date
    Mar 2005
    Modena, Italy
    Quote Originally Posted by BillCobb View Post
    OK, so what's your feeling about two other dynamic vehicle properties that are affected by the Mx's ?
    They felt pretty good.

  2. #82
    Hello everyone,
    I visualized for myself the jacking force. Maybe some of you find it at least “nice to look at”. I hope it is correct. If its looks wrong to you, please let me know...

    For this diagram, I have chosen a constant sum of lateral fore on the tires of 1000 N and a constant roll center height of 0,1 m above ground. So, the jacking force will switch from positive to negative values if you would place it under the ground. (The contact patch does not move and there are no moments MX included.)
    Because the jacking force is dependent of the slope of the n-lines and on the lateral force of each (left and right) tire, I varied the n-line slope by changing the lateral position of the RC (RC height is constant) and I also varied the distribution of lateral tire forces (left and right tire).
    Jacking Force.jpg


    a) If the RC height is zero or near zero you don’t have to deal with lateral roll center migration because there is no jacking.

    b) if your lateral RC position is (constantly) under one tire you will have a high jacking force. Here it will be better to have the lateral RC position under the inside wheel when the outside wheel is loaded. This will give little negative jacking (if the vertical RC position is above ground).

    c) If your lateral RC position is (constantly) under the loaded (outside) wheel, then you have the worst case.

    d) if your roll center travels lateral from infinity to zero and to the other side to infinity you probably won’t have to deal with the lateral position of the RC. Even if the RC height is large. Because:
    1.) the lateral movement of the RC is (I assume) very high. This means the time is very short when large jacking forces will occur (large jacking will occur if the lateral RC position is near the tire), thus there will occur only a small amount of lift on the vehicles body (because of the inertia of the body).
    2.) Because the lateral movement of the RC is load dependent, for large lateral RC movements there would only be a small amount of lateral tire force available at the lateral RC position near the tire (which are relatively small lateral movements).

    Cheers Jakob

  3. #83
    Senior Member
    Join Date
    Mar 2005
    Quote Originally Posted by _Jakob_ View Post
    For this diagram, I have chosen ... a constant roll center height of 0,1 m above ground...
    ... (The contact patch does not move and there are no moments MX included.)
    Because the jacking force is dependent of the slope of the n-lines ... I varied the n-line slope by changing the lateral position of the RC (RC height is constant [at 0.1 m above ground]).

    For clarity, let us further assume that this problem lies in the "Idealised 2-D Kinematics" world, where we only consider an end-view of the car and its suspension on a horizontal road, and where everything is perfectly rigid (so no floppy tyres), and where the wheels are thin enough that their "contact patches", namely the "wheelprints" that touch the ground, can be considered to be perfect geometrical points.

    Futhermore, we assume that each such Kinematically perfect suspension-linkage only allows its wheelprint to move along a single, unique "path of motion". That is, each suspension-linkage has only "1 DoF". In such a view, a wheelprint's "n-line" is the line that, at any instant, passes through the wheelprint, and is perpendicular (or "Normal") to the "path of motion" of the wheelprint. That is, the wheelprint can have NO movement, wrt the car-body, in the direction of this line.

    The above is the usual end-view picture presented in most books on this subject, just with the assumptions more clearly spelled out.

    IMPORTANT POINT - All realistic car suspensions have the "path of motion" of their wheelprints (wrt car-body!) running in a predominantly vertical manner. That is, the wheels bounce mostly "up and down", with only a lesser sideways movement. This means that the end-view "n-lines" MUST be predominantly horizontal. More bluntly, the n-lines are NEVER close to vertical. (Well, not until the car is half-way through a rollover!)

    And since the notional "RC" (in the sense it is used here), is the intersection point of the right and left wheelprint's n-lines, we can conclude that the RC can NEVER be directly above or below either wheelprint. Try drawing a RC directly above a wheelprint, while keeping that wheelprint's n-line at less than, say, 45 degrees from horizontal. Not possible!

    In fact, if you draw many positions of the RC as it migrates from under the centre of the car to some distance far to the left or right of the car, you will find that the RC's path ALWAYS PASSES THROUGH the wheelprints. So whenever it is at either wheelprint's lateral position, it MUST be at ground level, namely AT THE WHEELPRINT.

    To restress above, when you "...varied the n-line slope by changing the lateral position of the [constant height] RC" the result was to give the n-lines very UNREALISTIC slopes whenever the RC was close to either wheelprint.

    This is a good example of the misleading power of the RC. Thinking in terms of the RC first, lead you to select the unrealistic n-lines second. If you chose to think in terms of the n-lines first, say by varying them in a realistic range of +/- 45 degrees from horizontal, then the resulting RC positions would have been more realistic.

    But, don't feel too bad about making the above mistake. You are in good company. I recall one of Claude's seminar handouts (admittedly from a long time ago) that had exactly the same mistake. It showed the RC at about axle-height ... directly above a wheelprint!


    (PS. Also think about the RC when it is AT the left-wheelprint, which happens whenever the right-wheelprint's n-line is horizontal, while the left n-line is not. Now the left-wheelprint's n-line can be at any angle, and it can thus give any amount of +/- jacking, or anti- or pro-roll, depending also on the Fy force there. But any calculation that only uses the "RC-position" cannot determine these effects! This is because the RC-position's ability to determine the left-wheelprint's n-line-slope (aka its "control-arm-force-line-slope"), VANISHES when the RC merges with the wheelprint.)
    Last edited by Z; 04-14-2017 at 10:57 PM.

  4. #84
    Senior Member
    Join Date
    Mar 2005
    Since this thread has resurfaced I thought I might add some more on "gyro" issues.

    Back near the top of page 4 (September 2012) I discussed the gyroscopic-couples that act between the rapidly spinning components of a vehicle, such as its wheels and engine, and the rest of the vehicle, and asked,
    "Is it possible to lift the OUTSIDE wheels when cornering, say with a fast revving, reverse rotation, heavy flywheel?

    Moop answered,
    "...I was able to hack together a representation of a car that tried to take advantage of this phenomenon...
    It was quite a ridiculous car. A 20 cm diameter 50 cm long steel bar rotating at 4k RPM was not enough...
    ... worth a try though, haha."

    Indeed, a reverse-rotating engine/flywheel of typical size in FS/FSAE might be able to take the edge off the overturning roll-couple acting on the car during cornering (ie. the inward-wheel-Fy-forces and outward-centrifugal-force separated by CG-height), but it is unlikely to be able to "lift the OUTSIDE wheels". See my calcs that put numbers to this near the top of page 5 this thread (Sept 2012).

    However, I have recently had the pleasure of watching vehicles where such gyroscopic-couples are, in fact, the DOMINANT Vehicle-Dynamic forces. Yes, "BattleBots" has been airing somewhere up around Channel 93 (?) on the telly here lately. Most amusing!

    Here is "Blacksmith vs Minotaur" on YouTube.

    Minotaur is what they call a "drum-spinner", meaning it has a quite heavy toothed steel-bar that spins rapidly about a lateral axis. The bar spins such that its teeth move upwards at the front of the drum, giving an "upper-cut" punch to its opponent. So the drum "reverse-rotates" to the normal forwards direction of the wheels. I recall mention of the drum spinning at ~10,000 rpm, or higher?

    Some VD highlights:
    * Time = 0:19 -> 0:30 - Minotaur has had enough time to get its drum spinning at top speed, so whenever it moves quickly in yaw there is much "gyro-induced-roll". Looking in plan-view, with drum at front, a rapid clockwise movement causes Minotaur to lift its left-wheels. Rapid anti-clockwise yawing causes it to lift its right-wheels.

    * During some middle parts of the battle Minotaur's drum can be seen spinning quite slowly, due to impacts with its opponent. There is consequently much less lifting of left or right wheels during yawing motions.

    * ~3:20+ s - At the end of the battle Minotaur gets its drum back up to full speed, and does a "gyro-victory-dance"! Note that it has to move mostly in reverse to maintain its gravity defying roll-angle, because it only has "skid-steer" and not independently steered front-(or rear)-wheels. So, while driving in "reverse", with its drum thus spinning "forwards", it has an unusually large outwards gyro-roll-couple acting on the body (ie. in the direction that would be very bad for a racecar).

    Some things are very significant (= "important") in engineering design, while other things are not. Obviously, great care should be taken when estimating and calculating the significant things, but a "faster and looser" approach can be had with the insignificant ones.

    Roughly speaking (IMO), the "significance of things" can be split into three categories.

    * The 1%-ers, or less, = INSIGNIFICANT - These things have such a small influence on overall performance (ie. <1%) that they can be safely IGNORED. But, of course, this "very small influence" must first be very firmly established, before these things can be ignored. Ignoring these things can be helpful, because it gives more time to "fry the bigger fish".

    * The 100%-ers, = OVERWHELMING - These things are so obvious that no one misses them. It follows that these "first order" effects always appear at centre-stage in all considerations, calculations, and the like. For example, the many types of gyroscopic effects occurring in BattleBots means that even the commentators are well-versed in these phenomena. And any VD-analysis of Minotaur would be utterly useless if it did not include the gyroscopic efffects.

    Specifically, Minotaur is a very low and wide vehicle, with exceptionally low CG-height to track-width ratio. So, based on normal VD analysis, it should NOT "lift its wheels" during cornering. But it does! It even does so when yawing/spinning on the spot, when there is zero centrifugal force acting at the CG. This is because the dominant "rolling/jacking" forces acting on Minotaur are the gyroscopic-couples!

    * The 10%-ers, = SIGNIFICANT - This is the middle-ground, where things get tricky. These things have "smallish" effects, which, unfortunately, can be so small that they are overlooked. But often these effects are NOT SO SMALL that they can be safely ignored. The hard parts about these 10%-ers is, firstly, identifying them, and secondly, deciding how much effort to put into calculating their effects.

    Here I note that the gyroscopic effects of spinning wheels and engine parts on FS-cars are in the region of, or above, that 10% figure (as per the calcs back on page 5). So they are "significant".

    So..., how many of you include these gyroscopic effects in your VD-simulations?

    To restress why and where these gyro-couples are important, here is a clip from my 2012 post.

    "Now think about a driver sliding the tail around a hairpin so that the yaw velocity Wz is greater, and perhaps in first gear with the engine at 10,000 rpm, ... [giving] a much greater roll couple ... than normally calculated (maybe 150+%?)."

    My point is that an over-eager driver can easily spin their car by being too heavy on the accelerator coming out of slow corners. If the end result is that the gyro-roll-couple from a forward-rotating engine helps the car roll completely over, then ... game over for that Team. Even if the car only gets to the position of Minotaur during its victory-dance, then such a "dynamic response" from the car is likely to adversely affect the driver's confidence, and hence also their lap times.

    However, a reverse-rotating engine might provide a large enough gyro-roll-couple, acting in the right direction, that the driver can have complete confidence in the car. Regardless of how fast they "flick" the car around the hairpins, the car remains firmly planted on the road.

    End result = more driver confidence = faster lap times.

    Last edited by Z; 04-21-2017 at 09:02 PM.

  5. #85
    Hey Z,
    thank you very much for your detailed answer. What you said makes perfectly sense.
    It took me a while to reply to you...sorry for that. But there is nothing I can add to your post . So thank you very much again!
    Best regards

  6. #86

    The effect of gyro on jacking

    In reference to #40

    Hey Z, How are you? I really liked the way you explained the jacking forces in the 2D plane.
    I have a few questions.
    1. To increase the accuracy of the calculations I included jacking due to Mx, Gyro, Unsprung inertia along with the cornering force jacking. ( I assumed a fixed IC within the track)

    2. To calculate jacking due to Mx,
    I assumed a horizontal force acting at the wheel centre(assumed as the cg of the unsprung) in the central plane, which will provide the same moment at the assumed and fixed contact patch centre as provided by Mx.
    Now this force is resolved with parallelogram law on the n-line joining the wheel centre and the IC. Now, its jacking effect is calculated including the inertial force due to the unsprung mass as well.

    3. Then for the jacking force due to Gyro,

    First of all the couple on the whole vehicle due to Gyro divided by track gave me the load transfer which is added to the inertial load transfer(due to M*Ay) then I used the same method of n line slopes to calculate net jacking on sprung. Please correct me if i am wrong.

    4. OR I should also resolve the gyro couple (for each wheel, hub, disc and other rotating components) at the IC to get its additional jacking effect which reduces the roll gradient( for example- lets take the front right in a left hander, (SAE axis system) the gyro couple of this FR creates a moment in positive x direction about IC which would want to decrease the roll gradient of the car). This shows me that if somehow my Gyro couple increases, it decreases the roll gradient. Please correct me if I am wrong.

    But as I have written in (3) (assuming the same coefficient of friction of tyres) adding more gyro couple increases the load transfer where we get the same roll gradient which is similar to increasing the lateral accelaration to gain similar amount of increased load transfer as with gyro where the roll gradient remains the same(I repeat, I have assumed fixed IC and coefficient of friction).

    Anyone is welcome to answer.
    Please forgive me if any of my above thoughts are drastically differing from reality

    Thank you in advance!!

    Team Defianz Racing
    Delhi Technological University

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