Geo

That's where I'm having a hard time.

As for the issue of when the car is on the ground and trying to lift one side vs the other, don't forget that we are transfering the weight from our side of the car to the other. In that case, the springs on the other side are resisting us (or that's the way my monkey brain sees it).

Again, I'm not saying it's not true. Seems every other time I do that I find out I'm wrong and learn something new. If this case here is true, I'd like to understand it better.

Adam@Autometrics

Try this 1d analogy. Much easier to draw, but here goes...
on each side there is a vertical, imovable wall.
in between the walls you have a block mass in the center and a spring (horizontal) on each side of the block. So you have a block suspended between 2 walls by the 2 springs.

The springs have a free length of 6" and a rate of 200 lbs/in.

With no additional load added (assuming no friction) the springs are compressed to 4" in length, so each is exerting 400lbs on the block.

How much force will it take to move to block toward one wall 1"?
Now one spring will be compressed only 1" and will put 200lbs on the block.
The other spring is now compressed 3" and will put 600lbs on opposite side of the block.
So it will require 400lbs of force to move the block 1". Balance forces so that there is 600 lbs on each side of the block.

The loaded spring is 200lbs/in, but it takes 400lbs/in to compress it. That is because the spring that is becoming uncompressed is contributing to the movement resistance.

Now think of the same example, but you move the block 2". The inside spring becomes loose, and the outside spring is compressed 4" and will exert 800lbs on the block.

So, the movement resistance is still 400lbs/in.

Now move the block another 1" in the same direction, for 3" total movement, 5" total compression. This requires 1000lbs total force.
It takes only 200 additional lbs to move the block 1". The movement resistance is only 200lbs/in. This is a 50% loss in movement resistance once the uncompressed spring becomes completely unloaded.

I did this all linear, but it will work for rotation/roll exactly the same. When the inside spring becomes comletely unloaded, the roll resistance from the springs gets cut in half.

mds

Interesting Adam. So in a turn the inside spring, being compressed less, effectively adds roll resistance because its force is decreasing as it lengthens. So a dcreasing force on one side of the car is in a sense as effective as an increasing force on the other side of the car. Amazing.

Adam@Autometrics

EXACTLY! Thank you.
Apparently, my example wasn't completely abstract.

hrk

Very interesting.

So moving that block 1" from 4" to 3" requires 400 lbs.
Moving that block 1" from 3" to 2" requires additional 400 lbs on top of previous 400lbs, total 800.
But then when we go one more inch to 1" we require only additional 200 lbs totaling 1000 lbs.
Is this the reason for helper springs?
My head starts to hurt.

hrk

ehall

"Very interesting.

So moving that block 1" from 4" to 3" requires 400 lbs.
Moving that block 1" from 3" to 2" requires additional 400 lbs on top of previous 400lbs, total 800.
But then when we go one more inch to 1" we require only additional 200 lbs totaling 1000 lbs.
Is this the reason for helper springs?
My head starts to hurt. "

or are those tender springs?

Great thread. Thanks guys.

Jon Schepps

I just came across this thread, and the notion of a spring's unloading on one side adding to the spring on the other side seemed intuitively stange to me as well. Yet, Adam's simple "thought experiment" certainly shows this to be the case.

Nonetheless, it still bothered me. So in case anyone is still interested, I did the simple math. Take 2 identical springs of spring rate k, and lay them "nose-to-nose" between 2 walls such that both are compressed (with or without Adam's block between them). Let's call the compressed length of each spring, x. So in this case, the force each spring exerts on the other is simply kx=kx. External forces are zero.

Now apply an external force to compress one spring by an amount d (a small distance, such that the extending spring is not completely unloaded. The balanced force equation looks like:
k(x+d) = F + k(x-d)

(F can be on either side of the eq. depending whether you call it the external force or the restoring force). Now solve for F and you get:
F = 2kd

In other words, the effective spring rate doubled. F/d= 2k. With a single spring, Force/displacement is alway k.

Interesting. No doubt the extension to vertical springs supporting a horizontal platform (like our cars) acts similarly.

Jon.

996CAB

Waking this thread up - theory discussed here worked for me...see https://rennlist.com/forums/997-gt2-...des-etc-7.html