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12032009, 02:11 PM  #1  
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Motion ratios and damper piston velocity...how hard can it be?
Last edited by RR98ITR; 12042009 at 09:39 AM. 

12032009, 02:19 PM  #2 
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Re: Motion ratios and damper piston velocity...how hard can it be?

12032009, 04:39 PM  #3 
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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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12032009, 11:59 PM  #4  
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Re: Motion ratios and damper piston velocity...how hard can it be?
Quote:
By definition, the average velocity of shocks in use is also very close to zero.
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12042009, 07:16 AM  #5 
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Re: Motion ratios and damper piston velocity...how hard can it be?

12042009, 07:24 AM  #6  
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Re: Motion ratios and damper piston velocity...how hard can it be?
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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). 

12042009, 09:38 AM  #7 
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Re: Motion ratios and damper piston velocity...how hard can it be?

12042009, 10:01 AM  #8  
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Re: Motion ratios and damper piston velocity...how hard can it be?
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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) 

12042009, 02:43 PM  #9 
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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 coilover 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 coilover moves 0.75" (i.e. coilover 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 coilover 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 = coilover displacement (MR < 1) Wheel force/MR = coilover force coilover force/coilover displacement = (1/MR^2)*Wheel force/Wheel displacement or Wheel rate = MR^2*coilover rate. Now what is of course interesting is that the coilover 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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Tags 
coil, crx, damper, dampers, honda, laarms, motion, over, piston, ratio, ratios, shock, speed, spring, velocity 
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