suspension kinematic analysis

[Karam Atteia]

http://img801.imageshack.us/img801/913/sssba.png

i have a problem in the rear sus side view geometry .. as i 've designed for a 22% anti-squat , but after the analysis on SusProg3D .. the results made me confused as it gave me a -45% anti-squat . i attatched an image for the rear sus on SusProg to see.

[Claude Rouelle]

First things first ...
1. Do you know how to define anti-dive, anti-squat anti-lift? There are different way to quantify it. Which definition does Susprog use?
2. Do you know that the calculations are different if you have inboard or outboard brakes, solid axle or independent suspension?
3. Ideally those calculations are not only pure kinematics dependent: brake balance, torque distribution (LR / RR due to diff setting and front rear torque distribution if you have a 4 WD), in other words tire longitudinal forces have to be taken into account.

Let's see what your answer to question #1 is and we will go from there.

[Tim.Wright]

Well looking at that geometry I can see it will have negative anti-squat because your side view instant centre is below your wheel centre. Acceleration forces act forward at the wheel centre and will create an anticlockwise moment about the instant centre which puts the suspension into bump.

I think the problem is that in your calculations you are using the angle of the line from the contact patch to the instant centre. For anti squat (which is specific to the acceleration loadcase) you need to take the angle of the line from the wheel centre to the instant centre.

Tim

[Karam Atteia]

@ Claude
what i know that it is how much the sus. links are going to react the brake/acc. torque instead of the springs, but for no.2 i know the difference between the inboard and outboard brakes in the antix but i don't know the difference between the solid axle and the independent sus.
@ Tim
sry i can't get it .. why do the traction force act at the wheel centre .. it seems to me that it is like the difference bet. inboard/outboard brakes, isn't it ?? why ?

[Karam Atteia]

i saw that before in RCVD .. that for the independent sus. we take the line from the wheel center to the IC, but i can't understand it why.
but in tune to win it is not mentioned if there is a difference between dependent or indep. sus.
look at this , i think this have an independent sus.
http://img594.imageshack.us/img594/1858/123jjd.png

[Tim.Wright]

That looks wrong to me...

Do a free body diagram of the hub during braking and acceleration and it will become clearer.

Under acceleration, the only interface between the wheel and the upright is through the spindle bearing. This can transmit a force but not a torque. So a given force at the contact patch can only be reacted by an equal/opposite force at the spindle bearing.

Under braking you have two interfaces between the wheel and the upright. Through the spindle bearing (as before) and also the brake pad. The vector addition of the forces acting at these 2 points results in a force and a torque at the wheel centre. If you do the maths you will see that this wheel centre force + torque is mathematically equivalent to a single force (and no moment) acting at the contact patch.

Its a bit difficult to explain, but if you draw it up, it will become clear.

Tim

[Karam Atteia]

@ Tim
ok .. nw i get the point same as the inboard brakes there is no torque from the pads, and we brake from the shaft as we accelerate so we'll treat it as an inboard brake and take the line from the wheel centre.
put why there is a difference between the independent and dependent sus. in the squat

[Claude Rouelle]

Karam Atteia,

You really think this is correct?

I know this sketch comes form Carroll Smith, one of the person who really made FSAE what it is and one race car engineering author I avidly red all the books from. He was one of my inspiration when I did not know anything and there was no FSAE for the kid I was at that time. Ans Milliken RCVD did not exist. And no OptimumG like seminars. But today things are different. Knowledge has evolved in quality and quantity. And information circulates more quickly and is also more available. Your future automotive employers (if that is what you want to do) as well as vehicle dynamics & suspension FSAE design judges will want much, much, much more.

That definition
http://img594.imageshack.us/img594/1858/123jjd.png is over simplified.

1. Suppose you have a real bad rear tire with longitudinal friction coefficient of let's say Mux = 0.1 and one kind of soft compound F1 type with a Mux = 2.0. Do you think you anti-lift (braking) and anti-squat (acceleration) will be the same?

2. Also imagine that the brake distribution is 100 % front or 100 % rear, do you really think that the rear ride height (or teh front one) will dynamically changes the same way under the same input brake pedal force Vs time? Want an experiment? Pull the hand brake (if the hand brake acts on the rear wheel as it is for 99 % of today's car). Is the rear of the car suspended mass going up or going down? What does that tell you?

3. One another question. That is one of my favorite as a judge because it is simple and the answer puzzles the guys. But they learn from it. Let's say that when you brake at 100 km/h, the rear ride height increases from let's say 50 to 100 mm. Ignore any aero downforce. Now put rear springs let's say twice their original stiffness. Let's also say that the rear ride height has been adjusted so that at 100 km/h, just before braking it is also 50 mm. Let's say you brake pedal effort Vs time is exactly the same as the first time. You did not change any parts of the brakes, did not change the brake balance or anything else on the vehicle than just the rear springs. The question is: will you have the rear of the car moving up more or less than 100 mm?

Ii is not all about kinematics.

[Karam Atteia]

@ Claude
1. i think no, as the long. friction forces will not be the same . it is greater for the F1 type, so more dive/squat.

2. if all the brakes comes from the rear only as for using the hand-brake, i think the ride height will be decreasing (the car go down) till the car stops.and no front rise.
the amount of brake distribution affect the dive/rise.

3. i think i'll go to the 100 mm but in time faster than the before.

[Claude Rouelle]

Karam Atteia

For 1 and 2 You are right but if you were in design competition I would say "can you elaborate?"
At least you seem to understand that the anti-dive, anti-lift, anti-squat issues are not only kinematics dependent.

For 3, think again. Back to 100 mm? Why?

[Karam Atteia]

i think on it again .. i don't know if it is right .. if we have say 60 kg of load transferred to the front from the rear then in the first case if the spring say(k1=6 kg/mm) then the spring will move 10 mm , while in the second case and we still have the same amount of load transfer and if (k2= 12 kg/mm) then the spring will move 5 mm .

so, for ur example the car will move up to 75 mm

[Karam Atteia]

but i still don't get why there is a difference between the solid axle and independent sus. in the anti-squat geo.

[Warpspeed]

Originally posted by Karam Atteia:
but i still don't get why there is a difference between the solid axle and independent sus. in the anti-squat geo.
What matters is not if it is a beam axle or independent.
What matters are if the brakes are located inboard or outboard.

Think about what happens when the caliper grabs a firm hold of the disc.
With inboard brakes the brake torque is resisted directly by the whole sprung mass.
With outboard brakes the braking torque is resisted by the hub carrier, and it then gets fed directly into the suspension links.

The effect is very different.

[Karam Atteia]

@ tony
i'm talking about squat not dive

[Warpspeed]

Originally posted by Karam Atteia:
@ tony
i'm talking about squat not dive
O/k, same thing really in reverse.

With a live rear axle the torque reaction is carried by the axle housing directly into the suspension linkages.

With IRS, the torque reaction from whatever drives the diff goes direct into the sprung mass.

It's really a very similar but opposite situation to dive and braking.
There are two completely different scenarios in either case.

In one case the suspension linkages are completely isolated from any axle torque, and in the other the suspension linkages get twisted around the axle in the same direction when braking, and in the opposite direction when power is applied.

[Karam Atteia]

O/k, same thing really in reverse.

With a live rear axle the torque reaction is carried by the axle housing directly into the suspension linkages.

With IRS, the torque reaction from whatever drives the diff goes direct into the sprung mass.

It's really a very similar but opposite situation to dive and braking.
There are two completely different scenarios in either case.
i still can't understand it .. for inboard /outboard brakes i can feel the difference but for traction, i really can't feel any difference .
in braking really there is a torque generated by the pad which make the sus. links react it .
but what is the difference as the acc. torque is transmitted to the rear wheels by the same way for live axle or independent sus.

[Warpspeed]

Originally posted by Karam Atteia: torque is transmitted to the rear wheels by the same way for live axle or independent sus.
The torque is transmitted to the wheels the same way, but you cannot transmit torque from nothing.

Something somewhere has to have an equal and opposite torque reaction force to the generated axle torque.

The issue here is, does that torque reaction have to go through the suspension linkages, or does it feed direct into the sprung mass.

[Karam Atteia]

The issue here is, does that torque reaction have to go through the suspension linkages, or does it feed direct into the sprung mass.
and how can i know that . please more details

[PatClarke]

Karam,

You can know that by sitting down and working out for yourself where the forces go!

Stop looking AT the issue and start looking INTO the issue.

When understanding comes (it will!) all the mystery will be cleared and you will have learned something you will not forget.

If someone tells you it's 'Yeah, okay' and get on with life not having learned a thing!

Pat

[Karam Atteia]

You can know that by sitting down and working out for yourself where the forces go!
i've already did that more than a time and still can't get into the point .
okay i'll try it again
thanks for ur attention all of you.

[Edward M. Kasprzak]

Originally posted by Karam Atteia:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">You can know that by sitting down and working out for yourself where the forces go!
i've already did that more than a time and still can't get into the point .
okay i'll try it again
thanks for ur attention all of you. </div></BLOCKQUOTE>

Don't just "try it again"! This thread has given you some very good direction. It's time to sit down with your Faculty Advisor (or some other professor, graduate student, upper-classman, etc.) who can help you put the pieces together. A few minutes face-to-face someone who understands free body diagrams should do it. Make some sketches and work-out where the forces and torques are reacted.

[M. Nader]

Hi Karam, If i understand correctly you are from Egypt.

I think i can help you with the problem you have, if you are yet to find a solution. i am only free this weekend though http://fsae.com/groupee_common/emoticons/icon_smile.gif

Send me a private message and we can work it out from there