Jacking force

[Flight909]

Thanks you for this post Z.

I agree with b,c,d but I think a is different. You make a assumption of the distribution of the lateral component of Fl and Fr. In reality (with "real" tires) this cannot be determined in this geometric way. Do you agree?

I have to say that 1,2,3 Axioms was covered in my classes, but maybe not enough practise to use them.

[Z]

Originally posted by Flight909:
I agree with b,c,d but I think a is different. You make a assumption of the distribution of the lateral component of Fl and Fr. In reality (with "real" tires) this cannot be determined in this geometric way. Do you agree?

Flight909,

Did I have to assume the "distribution of the lateral components of Fl and Fr"? Not really.

If given Fg and Fi, and the CG and wheelprint positions (and toe-angles, see below), then I might only assume the "Coefficient of Friction" of one wheelprint. This gives the slope of its LoA, which then passes through Fg+i's LoA, and all the other information is found.

Note that by assuming this intersection point of the wheelprint LoA and the Fg+i LoA, the magnitudes of the wheelprint forces are immediately determined, and hence so also is the distribution of their lateral components. So it could be said that the Fy distribution is either "assumed", or "solved for", SIMULTANEOUSLY with all other unknowns.

If you have good tyre data, then a sensible approach might be to assume that the two wheels have CFs in a certain ratio. So with normal Tyre Load Sensitivity the heavily loaded outer wheel might have, say, 10% less CF than the lightly loaded inner wheel. The two wheelprint forces' LoAs now intersect below the car, and all information is found. If both wheels have equal CFs, then wheelprint LoAs and Fg+i are all parallel, and again all information is found. (Note that in the sketch the right (inner) wheel has an unrealistically high CF (ie. too close to horizontal), but that was to squeeze it into the space available.)
~~~o0o~~~

Both geometric and algebraic approaches can only "solve" for the same number of unknowns. I prefer the geometric approach because as a map it is much closer to the terrain.

So at one extreme you might have all the information (from strain gauges, accelerometers, etc.) and the geometric approach just gives a picture of what is already known. But this picture gives a better understanding, IMO, than, say, a table of numbers. At the other extreme you might be assuming ALL the information (for "big picture" exploration), and again pictures give better understanding, IMO.
~~~o0o~~~

It is worth pointing out that a car can have "jacking" even when running in a straight line on a level road. The lateral Fy tyre forces are very sensitive to toe-angles. So if the car has an above ground RC (as in sketch), and the wheels are toed-in, then inward Fy forces act at each wheelprint, and the body is jacked up.

This is easily demonstrated by strapping on some roller-skates, or ice-skates. Get some speed up in a straight line, and move your feet wide apart so that your legs represent the n-lines. Toe-in and your body "jacks up". Toe-out and your body "jacks down"... Ouch!!!
~~~o0o~~~

The original VW Beetle is perhaps the best known car to suffer from "interesting" handling as a result of steep n-lines and jacking. The diff-centred swing-axles give n-line slopes of about 1:2 (= rise:run, = Fjacking:Fy). However, the swing-axle is "semi-trailing" (pivot at ~45 degrees to car centreline), and this gives increasingly large toe-in as the wheels move away from ride height.

So when cornering the greater jacking force at the outside wheel lifts the body up a bit. This adds a small amount of toe-in, which lifts the body a bit more. This increases the n-line slopes, lifting the body yet another bit. But now the toe-in is increasing at an increasing rate... The tail can jump upwards so much that both rear wheels leave the ground. It's a bit like squeezing a pumpkin seed between your fingers...

Z

PS. I note that newbies always ask for "equations". The more experienced FSAEers tell them to start with a FBD. I reckon stay with the FBD, and only use equations when you want answers to five significant figures or more (the last four of which are probably wrong... ).

[Z]

GYROSCOPIC ROLL COUPLES.
=========================
The following might be useful to anyone wanting to do moderately accurate Vehicle Dynamics simulations. These are simple calculations of the gyroscopic roll couples caused by spinning wheels, engines, or flywheels, during cornering.

(BTW, does anyone already include these gyro effects in their VD sims?)
~~~o0o~~~

CAR CORNERING ROLL COUPLE.
=============================
Firstly, as a baseline, the roll couple on the car due to the centrifugal inertial force (Fi) acting through its CG, and the centripetal wheelprint forces (Fy) acting at ground level is,

Tcar = Mc.Ay.Hcg

where,
Tcar = the roll couple acting on the car (Nm),
Mc = total mass of car (kg),
Ay = lateral (cornering) acceleration of car (m/s.s)
Hcg = height of CG (m).

Putting in some numbers for, say, a medium weight Mc=250kg car+driver, cornering at Ay=~1.2G, and with lowish CG height of Hcg=0.25m, gives,

Tcar = 250 x 12 x 0.25 = 750 Nm.
~~~o0o~~~

Gyroscopic roll couples can be found from this simplification of Euler's Rigid Body Equations (*see note below),

Tx = Iy.Wy.Wz

where,
Tx = the gyroscopic roll couple of the spinning body (eg. wheel) about the longitudinal X axis, (Nm),
Iy = the 2nd Mass-MoI of the body about the lateral Y axis, (kg.m.m),
Wy = rotating velocity of the body about the lateral Y axis, (radians/s),
Wz = yawing velocity of the body about the vertical Z axis, (rad/s).
~~~o0o~~~

WHEELS.
=========
Now consider the gyro roll couple of the four wheels of radius Rw=0.25m, as the car goes around a hairpin, taking the average corner radius for all wheels Rc=3m, again at Ay=1.2G.

Assume each wheel has rotating mass Mw=10kg, at radius of gyration Rg=0.2m, for Iwy= Mw.Rg.Rg = 0.4 kg.m.m.
(This is less than a 13" steel road wheel & tyre, but more than a 10" FSAE carbonfibre wheel & racing slick.)

Average forward velocity of car (and also wheels) Vx = sqrt(Ay.Rc) = sqrt(12 x 3) = 6 m/s.

Average rotating velocity of wheels Wwy = Vx/Rw = 6/0.25 = 24 rad/s.

Yawing velocity of car (and also wheels) Wz = sqrt(Ay/Rc) = sqrt(12/3) = 2 rad/s.

Twheels = 4 x 0.4 x 24 x 2 = 77 Nm.

This is slightly more than 10% of Tcar above, so quite significant. The direction of this couple is to roll the car outwards, so it adds to Tcar.
~~~o0o~~~

ENGINE.
========
Now consider the gyro roll couple from a lateral crank (East-West) engine.

Assume a rotating mass (crank, flywheel, clutch, etc.) = 10kg, at radius of gyration = 0.1m, for Iey = 0.1 kg.m.m.
(Perhaps similar to the very heavy crank+ of a Jawa single, although most FSAE engines would be less.)

Assume (???) rotational velocity of 5,000 rpm, or Wey = ~500 rad/sec (because Pi = ~3 ).

Yawing velocity of car/engine as before Wz = 2 rad/s.

Tengine = 0.1 x 500 x 2 = 100 Nm.

This is ~13% of Tcar, which is even more significant. This couple is added to the above car and wheel couples for a forward rotating engine (so giving 123% of the "standard" car-only calculation), and subtracted for reverse rotation (= 97% of standard calc).

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 10,000rpm. A car with forward rotating engine and heavy flywheel now has a much greater roll couple acting on it, than that which is normally calculated (maybe 150+%?), and may lift the inside wheels. Conversely, the gyro couple from a reverse rotating engine, or energy storage flywheel, counters the normal car+wheels roll couple, so helps keep the inside wheels planted.
~~~o0o~~~

It is interesting to look at the ratios of wheel and engine gyro roll couples, to the car-only roll couple.

1. Twheels/Tcar = 4.Iwy/(Mc.Hcg.Rw), or roughly = ~4.Mw.Rw/(Mc.Hcg).

This ratio depends only on the mass and geometry of the wheels and car (assuming that the wheels roll with no slipping). So in the above examples the wheels' gyro roll couple will always be 10% of the car-only roll couple, regardless of corner radius or lateral Gs.

2. Tengine/Tcar = Iey.Wey/(Mc.Hcg.sqrt(Ay.Rc)).

Here the ratio varies with engine speed (and so also gear), and cornering speed (since sqrt(Ay.Rc)=Vx).

In the numerical examples above (ie. 10% & 13%) the gyro roll couples are relatively high, mainly because FSAE cars are relatively small, compared with their wheels and engines. That is, bigger, heavier cars with relatively smaller, lighter wheels and engines, are not so affected by the gyroscopic couples (well, except perhaps when they spin...).
~~~o0o~~~

* As should be obvious, the T=IWW equation only works for spinning bodies that are "dynamically balanced" (ie. they spin about a "principal axis"). But is that enough?

For example, does a dynamically balanced two-bladed aeroplane propellor have the same gyroscopic behaviour as three-or-more-bladed propellors?

The answer is ... no! The two-bladed propellor has a gyro couple that varies from zero to double that given by the equation, twice each revolution (ie. horrible vibration). Euler's Equations, or a consideration of the "moment of momentum" of each blade, explains why...

Z

[formula1_dream]

Hi,
Thank you all for such valuable posts. It took me a while to get hold of the concept, but I think I have a good enough idea about the jacking force concept.

@ 'Z': too good explanation (Respect )

Regards,
Varun

[moksharmywebber]

Thank u very much "Z" for your valuable information...but there still few questions about "N-lines" in our mind which we would like to ask you.
--------------------------------------------------
Sir, u said -
"N-lines" are simply the directions in which a particular linkage (in this case the suspension control arms) can carry forces without the linkage moving. IMPORTANTLY, any forces that do not lie along n-lines must be carried by some other structure, which in the case of suspension is the spring-dampers, ARBs, etc.


(BTW, the concept of n-lines (= "normal-lines", aka "right-lines", because at right-angles to direction of motion) was developed in the 1800s. The lines drawn through the "pin-ended-links" in the 2-D sketch used to find the IC are n-lines (because "no" motion possible along them). A "contact normal" between a cam and follower, or between two gear teeth, etc., is also an n-line. They are everywhere, and are very useful simplifying tools.)"
--------------------------------------------------
1.Why is the line from tire contact patch to instantaneous center called a "n-line" ?

2.How can we say that forces from control arms are directed along "n-line" only?

Thanking you.

[Z]

moksharmywebber,

1. "Why is the line from tire contact patch to instantaneous center called a "n-line" ?"

Assume that we are considering a 2-D sketch of a "four-bar" linkage, as is typical of a front-view of a double-wishbone suspension. The top and bottom links represent the top and bottom wishbones. The left link represents the upright, and the right link represents the chassis.

Assume the wishbones converge towards the right, so lines through their ball-joints converge at an "Instant Centre" at some distance to the right.

If we assume that the chassis is "fixed", then the upright can only move "at the instant" by rotating about the IC. That means that during movement, all points on the upright describe arcs with centres at the IC.

The motions of all points on the upright are thus "normal" to (or at "right-angles" to) lines that radiate outwards from the IC. Thus ALL RADIAL LINES passing through the IC are "n-lines".
~~~o0o~~~

2. "How can we say that forces from control arms are directed along "n-line" only?'

Again consider the above four-bar linkage. Better yet make one out of four bits of cardboard, pinned together at the link-joints with thumbtacks, or similar.

Now "fix" the chassis-link (pin it to the table), and apply a force (in the plane) acting along a LoA that passes through any point on the upright-link.

If the force acting on the upright is NOT directed along its abovementioned n-line, then the force will cause the linkage to move (ie. wishbones will rotate about chassis joints), without the links carrying any load. ONLY when the force is directed along its n-line (ie. the force's LoA passes through the IC) do the links become stressed, and no longer move.

If you add a spring to resist up/down movement of the upright, then any force acting on the upright can be decomposed into two forces. One force with LoA along the n-line, which stresses the wishbone links. And a second force with LoA along the spring, which causes it to defrom.
~~~o0o~~~

The demonstration with bits of cardboard is much easier to understand than these words. Please try it.

Z

[Marshall Grice]

anyone notice this thread spawned an article in the latest issue of racecar engineering (dec'12).

[moksharmywebber]

Thank you very much Z sir... the demonstration helped us!
It is our pleasure to learn from you and get our concepts cleared !!

[Sid@TR]

Z,

our team has made a file regarding what we have interpreted and understood about roll-center and transmission of forces from TCP to the sprung mass-N-line concept.

http://www.scribd.com/doc/112763285/N-Line-Concept

Sir, please find time to have a look at it!
Eagerly waiting for your comments and suggestions!

[Z]

Originally posted by Marshall Grice:
anyone notice this thread spawned an article in the latest issue of racecar engineering (dec'12).

Marshall,

My brief look at that article on the RCE website convinces me that cancelling my subscription was the right decision. Also that the education system has gone further down the crapper than I ever imagined...

I am finding it difficult to express my feelings. Err..., maybe ... grrrrrrrr... and a few more not suitable here...

Why the editors print such nonsense is beyond me (because racecar?), but here are three reasons for rejection.

1. Very sloppy.
In Figure 3 Danny shows a FBD of the upright that is NOT in equilibrium. The missing force is the "jacking force". In words he describes this force as "the force applied straight to the tyre", and "directly into the tyre". This ambiguity suggests that he has little understanding of where or how this force acts.

2. Factually wrong.
Once again Danny claims that lateral tyre forces MUST be moved to his "FAPs" for correct results, and calculations done at the RC, or anywhere else, will give incorrect results. Bulldust! FWIW, this is probably because Danny does not yet realise that he must ALSO move the jacking forces to whichever point he is doing his calculations. The fact that FAPs are on a vertical line through the CG means that the vertical jacking forces exert no moment about the CG. So forgetting to move the jacking forces to the FAPs "accidently" gives the right answers, but I doubt he understands that.

3. Disgracefully slanderous.
Danny refers several times to "d'Alembert's approximation", which "... runs out of steam" for these suspension calculations. Huh???!!! Since d'Alemberts's PRINCIPLE (!!!) was proposed 250+ years ago it has always given EXACTLY the same results as Newton's Laws of Motion. Hardly surprising, given that it is simply a slightly different way of writing N2 (ie. instead of F=Pdot (F=ma) in dynamics, write F-Pdot=0 and solve as in statics). But now Danny reckons Newton's and d'Alembert's works are "approximations", and, fortunately for us, he will set us straight??? (I'm feeling ill...)

How can any of this pass as engineering? Motorsport, maybe ... but ...

I can't go on...
~~~o0o~~~

Sid,

I had a quick look at your link. Mostly right, but I will comment further when I climb down off the walls.

Z