## Dampers bump - rebound

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mep
34
Joined: Sat Oct 11, 2003 2:48 pm
Location: Germany

### Dampers bump - rebound

I am playing F1 Challenge 99-02 somethimes to
test new tracks and setups and there is one thing I wonder about
quite a long time now.

It's possible to setup the dampers at,
fast bump
fast rebound

slow bump
slow rebound
at front and rear tires. (Messured at N/mm/s)

But it is possible to make the rebound much more harder than the bump,
and at setups from others I see that some even do that.

I don't understand what the theory is, of having the rebound slower than
the bump especially for fast bumps.

For example you drive over a little hill in the road, your tire is forced
upwards with a strong force and your spring is compressed.
Now after the hill your tire has no ground contact and falls back
due to gravity and with a little force from the spring but the fall is
braked because you have the rebound dampers on a high level
(slow- much N/mm/s).
So you should loose grip.

So what is the sence behind it?

And how does this look like for slow bumps/rebounds?

vyselegend
0
Joined: Mon Feb 20, 2006 4:05 pm
Location: Paris, France

Playing with the bump / rebound properties wildely for sure modified car behavior, but I wasn't able to draw any logical principle from this, as the car can cumulate oversteer and understeer in the same run, and as I've no realistic telemetry anyway, I can't even tell for sure if the car bottomed or not, although it's most likely the only explaination left when you suddenly spin on a fast turn (uphill especially).

mep
34
Joined: Sat Oct 11, 2003 2:48 pm
Location: Germany
As for fast rebound I have it as low as possible and the bump
a bit higher (slower) because I think the tire should fall back as fast as possible.

But for slow bump/rebound I have no idear.

mep
34
Joined: Sat Oct 11, 2003 2:48 pm
Location: Germany
vyselegend
I can't even tell for sure if the car bottomed
For bottoming out I have another question:
Even when I drive with the lowest ground clearance
do I only have ground contact when driving over a smal
wave or hill in the surface instead of having permanent ground
contact at high speed.
Same goes for ground distance?

But I don't wan't to annoy anybody for asking special thinks to the game.
So lets keep on the damper thing.

MildaEvilda
MildaEvilda
0
Joined: Thu Mar 08, 2007 10:07 pm
Location: Czech Republic
You must take another variable to your calculations. The dynamic of the car that is pushed down by its weight. It may not be an issue when we talk about fast bounds/rebounds but slow ones are highly affected by the car dynamic.
When the car goes over the hill the tires are pushed up and car is pushed up too, especially with stiff suspension setup. When the tires are pushed down too quickly and strongly, the car has no time to drop down and it's bounced up again by the tires hitting the ground.

RH1300S
RH1300S
1
Joined: Mon Jun 06, 2005 2:29 pm

When a car moves in bump, the wheel accelerates against the force of the spring and damper. The damping force is there to control the movement of the wheel. In it's most basic form - to stop the wheel accelerating up off a bump so fast that it leaves the road.

When the wheel moves in rebound the damping is trying to control the motion of the wheel returning to it's original position while the spring is trying to pushing it back as fast as possible.

To my simple mind - you need a bigger damping force to control the rebound side than the bump side.

In practical terms for a race car (again very simplistic). The bump damping helps control how fast a car takes a set in a corner. The rebound damping helps contain that energy when the spring is trying to push the car body back up.

If the rebound was too weak, you would turn into a corner and the car would keep coming back to itself and not maintain a steady state. Too much rebound damping (in extremes) can make a car "pack down" - i.e. each time the spring is compressed, the damping does to realease the car quickly enough to resume it's original position. So at the next upward wheel movement the wheel starts with less travel to play with.

mep
34
Joined: Sat Oct 11, 2003 2:48 pm
Location: Germany
To prevent misunderstandings with little hills I mean smal irregularities
in the road surface maybe 2 -4 cm high.

RH1300S
If I got you right, you also mean that rebound should be
faster than bump because if not it could be
possible that the spring gets more and more compressed and
can't decompress.

MildaEvilda
Yor idear is quite good, this gives the car so a kind of bouncing
with hard springs.
But I have read in a setup guide that on tracks with many
shikanes where you have to drive over the curbs you use
harder springs.
But this is inexplicable with this theory.

Anyway you bring me to another thing to discuss about and that is going
to really get difficult now.
What happens exactly when you drive over a smal irregularity
maybe 3 cm high?
Does your wheel go up because it has a smal mass and is easier
to accelerate but pushed down by the spring.
Or does your wheel stay rigid to the car and the hole car,
which has a big mass, goes up?

It's clear it will be a mix between this but can you calculate
how much it will be?
Imagine you have a given car speed, spring stiffness, car weight,
tire weight and so on.
Can somebody put this into a physical formel to calculate
the amount the spring is compressed.
I tried this once but failed immediately.

MildaEvilda
MildaEvilda
0
Joined: Thu Mar 08, 2007 10:07 pm
Location: Czech Republic
mep wrote:
It's clear it will be a mix between this but can you calculate
how much it will be?
Imagine you have a given car speed, spring stiffness, car weight,
tire weight and so on.
Can somebody put this into a physical formel to calculate
the amount the spring is compressed.
I tried this once but failed immediately.
I can make the mathematic model, it's no problem but I don't know if we can figure right constants.

Ciro Pabón
218
Joined: Tue May 10, 2005 11:31 pm
mep wrote:... Can somebody put this into a physical formel to calculate the amount the spring is compressed.
I tried this once but failed immediately.
I can give you the simplified form civil engineers use to calculate the IRI (International Roughness Index) wich measures how many millimeters you move up and down while you move one millimeter along a road: it's the old and well known "Quarter Car Model". I don't have at hand the equations (I made once a program to solve the integral you have to make, but I was young and did not have children asking me to play outside, like today ). Perhaps someone can give you his own program or you can google for it. There is also a "Half Car Model" we discussed at some thread here, more complicated, which is more accurate if you want to reproduce the pitch sensitivity of the aerodynamic components of the car.

This is the model:

The car chassis is the "sprung mass" ms.

The wheel assembly is the "unsprung mass" mus.

The spring is ks and the damper is bs, placed between the car body and the wheel assembly

The spring kt is the tyre.

The variables xs, xus, and r are the car body travel, the wheel travel, and the road disturbance or bump.

The force fs, kN, applied between the sprung and unsprung masses, is controlled by feedback and represents the active component of the suspension system. If your car has not active suspension, omit it.

In that condition (no active suspension) if you define x1:=Xs', x2 = d(Xs), x3:=Xus and x4:=d(Xus), were d() means the derivative, then this are the equations that describe the behaviour of the quarter car dynamics (the dot on top of variables means the derivative):

Typical values for IRI calculation are:

ms = 290 kg
mus = 59 kg
bs = 1000 N/m/s
ks = 16182 N/m
kt = 190000 N/m

Hope this helps. If I have time (sorry, I have to play football with my 8 years old kid AND he is tugging my arm while I write this! ) I'll check my old class notes to look for the solution to the equations.

What you're looking for is the difference between X1 and X3 (or Xs and Xus) wich shall give you the spring compression, BTW. The solution I have somewhere buried gives you only X1, that is, how much the chassis moves up and down.

You have to integrate to solve the differential equations. I'm sure someone has to have the equation solved. Perhaps Reca? Or you can use Mathematica, if you have it.
Ciro

mep
34
Joined: Sat Oct 11, 2003 2:48 pm
Location: Germany
thank you ciro
it's not so importand better play with your son.
The damper think is more interresting.

David Anthony
David Anthony
0
Joined: Wed Feb 21, 2007 2:24 pm
mep if u ever step over to rfactor, and wanna race in a leauge pm me. I race for http://www.venezuelaf1.net/ its a international race leauge. We do all kinds of this tuning stuff, dampers etc etc. everything counts to win a race here. LOL

With damper i usualy have them always soft with the rebound a little stiffer. The fast dampers are generaly used for realy fast movements, like riding curbs, for exmaple monza, the first turn, u stiffen the fast dampers as it helps stable the car when you go over the curbs.

It also makes the car driver smoother at high speeds, if u are getting bouncing at high speed play around witht them, and the 3rd spring.
http://www.f1torque.com - Formula 1 chat - Talk the Torque

mep
34
Joined: Sat Oct 11, 2003 2:48 pm
Location: Germany
Last time an ingeneer told me that the rebound is slower
because of the springs bouncing back and so the tyres losing ground
contact.
This would make sence for the third spring making rebound stiffer
than bump.

But what about the two outer springs they are in some case
conected with roll so would it make sence to make rebound faster
then bump so that the chasis comes fast back to normal position
after driving through a corner?

modbaraban
0
Joined: Thu Apr 05, 2007 4:44 pm
Location: Kyiv, Ukraine
This may clarify it

mep
34
Joined: Sat Oct 11, 2003 2:48 pm
Location: Germany
Thanks for the link but I have a problem with getting the car low to the ground.
Packers do the oposit.
I have the theory that the car has always to be lower at front than
at rear.
But I didn't get the front lower. And at rear I have suspension as soft
as possible for ideal corner behavior.

modbaraban
0
Joined: Thu Apr 05, 2007 4:44 pm
Location: Kyiv, Ukraine
mep wrote:I have the theory that the car has always to be lower at front than at rear.
Sure! In my setups the rear is almost twice as high as the front!