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Old 12-03-2009, 03:11 PM   #1
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Default Motion ratios and damper piston velocity...how hard can it be?

I was just thinking about this...for no particular reason...

Imagine some people, trying to figure out if motion ratio mattered in the determination of how much force you wanted to get at what damper piston velocity...and they are making it a Hard problem...it shouldn't be That hard...

So imagine you had a really good motion ratio of 1 to 1...the piston velocity would match the wheel/input velocity...distance divided by time.

Then imagine you had a lesser motion ratio of 0.5 to 1 (the damper piston moves only 1/2 as far as the wheel)...Hmmmm...so if the damper piston only travels 1/2 the distance in the same unit time, then it's velocity would be 1/2...at all times thru the event.

** Ok, there is just a little bit more to it...that thing about motion ratio squared...K sub s times MR squared equals K sub w...

Imagine three cars otherwise identical save for their motion ratios...one is 1.0 , one 0.75 and the other 0.50...and a linear damper making 100 lbs force at 4 inches per second on the dyno...and a car at a point on the track where the wheel/input velocity (peak/average/whatever) is 4 inches per second...at MR of 1.0 the force exerted by the damper at the wheel is 100 lbs...at MR of 0.75 the velocity is 3 inches per second and the force is 75 lbs at the damper, and 42 lbs at the wheel...and at MR of 0.50 the velocity is 2 inches per second and the force is 50 lbs at the damper, and 12.5 lbs at the wheel. ** (ERROR - see post below)

In short the motion ratio matters - that's why your shock guy (or gal) asks for it.

When people talk about velocity generally, the unspoken assumption is that the MR is 1.0 - otherwise who knows what you're talking about. If that makes you wonder how many people know what they're talking about, or if they're taking proper care in their terms, you might be onto something. ** (Look out, you might indict your own self!)

Scott, whose shocks average velocity is very close to zero and still converging steadily...

Last edited by RR98ITR; 12-04-2009 at 10:39 AM.
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Old 12-03-2009, 03:19 PM   #2
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Default Re: Motion ratios and damper piston velocity...how hard can it be?

Quote:
Originally Posted by RR98ITR View Post
I was just thinking about this...for no particular reason...
hahha don't believe it.

BTW i like the way you explain it. It makes sense to me know.
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Old 12-03-2009, 05:39 PM   #3
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Default Re: Motion ratios and damper piston velocity...how hard can it be?

To which one could add % of critical damping, both at high and low velocity (damper shaft that is), for bump and rebound.
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Old 12-04-2009, 12:59 AM   #4
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Default Re: Motion ratios and damper piston velocity...how hard can it be?

Quote:
Originally Posted by RR98ITR View Post
...at MR of 0.75 the velocity is 3 inches per second and the force is 75 lbs at the damper, and 42 lbs at the wheel...and at MR of 0.50 the velocity is 2 inches per second and the force is 50 lbs at the damper, and 12.5 lbs at the wheel.
Try again...

Quote:
Originally Posted by RR98ITR View Post
Scott, whose shocks average velocity is very close to zero and still converging steadily...
By definition, the average velocity of shocks in use is also very close to zero.
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Old 12-04-2009, 08:16 AM   #5
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Default Re: Motion ratios and damper piston velocity...how hard can it be?

Quote:
Originally Posted by RR98ITR View Post
I was just thinking about this...for no particular reason...
SPY!!!!!!!!!
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Old 12-04-2009, 08:24 AM   #6
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Default Re: Motion ratios and damper piston velocity...how hard can it be?

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Originally Posted by beanbag View Post
Try again...



By definition, the average velocity of shocks in use is also very close to zero.


He's right.

Tangential Velocity at the ball joint (assuming MR of 1) = 4in/sec

Angular Velocity (constant along the entire rigid control arm) = TanVel/Radius or 4/1

So W = 4 rad/s

W = V/R

4 = V/MR (assuming that total length of arm = 1 for calculation's sake)

So for MR of .75, V = 3; MR of .5, V = 2 etc.



Also - the forces are correct as well. Only RATE uses the MR squared. Forces are directly proportional to the MR (ignoring angle of spring/shock for simplicity).
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Old 12-04-2009, 10:38 AM   #7
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Default Re: Motion ratios and damper piston velocity...how hard can it be?

Quote:
Originally Posted by beanbag View Post
Try again...
Uh-oh...inappropriate use of the MR^2 thing..."Inconceivable!"....

So then, that gives forces at the wheel of 25lbs for MR of 0.5, and 56lbs for MR of 0.75.

Scott, who offers his apology and his thanks...
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Old 12-04-2009, 11:01 AM   #8
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Default Re: Motion ratios and damper piston velocity...how hard can it be?

Quote:
Originally Posted by RR98ITR View Post
Uh-oh...inappropriate use of the MR^2 thing..."Inconceivable!"....

So then, that gives forces at the wheel of 25lbs for MR of 0.5, and 56lbs for MR of 0.75.

Scott, who offers his apology and his thanks...


You were right the first time Scott.

Summation of moments around the inner pivot point

F1 = force at outer ball joint
F2 = force at shock pivot point (opposite direction of F1)
D1 = Distance from inner pivot point to shock pivot point
D2 = Distance from inner pivot point to outer ball joint

F1*D2 = F2*D1

F1 = F2*D1/D2

if we considered angle of spring/shock it would actually be:

F1*D2 = F2*D1*sin(theta)

where theta is angle between spring/shock and control arm (<45deg)
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Old 12-04-2009, 03:43 PM   #9
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Default Re: Motion ratios and damper piston velocity...how hard can it be?

You guys are confusing me. Right, not right, right, not right!

MR^2 is for rates.
MR is for motion (displacements) and forces. No need for angles as you don't need to know lengths and displacements, just displacements. Measure wheel displacement and coil-over displacement.

Say wheel moves up 1" in 1/4 second (4 in/s velocity) and is resisted by a force of say 100 lbs. So force at the wheel is 100 lbs and wheel rate is 100lbs for 1" of travel or 100 lbs/in. If motion ratio is 0.75, that mean by definition that when the wheel moves 1", the coil-over moves 0.75" (i.e. coil-over displacement = MR*wheel displacement).

Since there is a lever causing this change in displacement, there is also a proportionate change in the force by the same but inverse ratio. For every fraction the displacement is reduced, the force is increased. Thus if the force exerted at the wheel is 100 lbs, then the force exerted on the damper is the wheel force divided by MR, and is thus 100lbs/0.75 or 133.3 lbs. The spring rate at the coil-over is thus 133.3 lbs/0.75" displacement = 177.77 lbs/in. This is the same as wheel rate/MR^2 = 100/(0.75^2) = 177.77 lbs/in.

So:

Wheel displacement*MR = coil-over displacement (MR < 1)
Wheel force/MR = coil-over force
coil-over force/coil-over displacement = (1/MR^2)*Wheel force/Wheel displacement

or Wheel rate = MR^2*coil-over rate.

Now what is of course interesting is that the coil-over velocity will be much less than the wheel velocity, and this is often overlooked. If wheel moves at 4"/s, then damper moves at .75*4"/s or 3"/s. If damper is linear and makes a force of 100 lbs at 4"/s, then it will make at force of 75 lbs at 3"/s. The force at the wheel is MR times this or 0.75*75 lbs which is 56.25 lbs. This is a lot less than you might have expected if you had assumed the damper velocity was the same as the wheel velocity. What's an MR between friends?
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