Spring rate change and torsion bar ?
10-12-2009, 12:14 AM
#16
Not sure why you are using 0.65 for the rear spring rate, it should be 0.42 And also, I would like a ratio that is a bit lower. That is a good reason to replace the front struts with perch mount struts with can use common springs. It is hard to find springs over 250lb for a stock strut.
Last edited by Bri Bro; 10-12-2009 at 12:33 AM.
10-12-2009, 07:03 PM
#17
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Yes, in stock configuration, the front a rear as close to the same spring rate. As you increase the spring rate, most folks go towards a stiffer front spring rate. In stock configuration you also get rear end squat on acceleration and a lot of body roll in the front end.
Not sure why you are using 0.65 for the rear spring rate, it should be 0.42 And also, I would like a ratio that is a bit lower. That is a good reason to replace the front struts with perch mount struts with can use common springs. It is hard to find springs over 250lb for a stock strut.
Not sure why you are using 0.65 for the rear spring rate, it should be 0.42 And also, I would like a ratio that is a bit lower. That is a good reason to replace the front struts with perch mount struts with can use common springs. It is hard to find springs over 250lb for a stock strut.
it's down now
http://www.tech-session.com/kb/index...x_v2&id=69&c=4
but the equation is:
TB rate + 0.65 x coil rate = overall effective rate
same equation is referenced here http://dsims.myzen.co.uk/944t/suspension.htm
Increase rear spring rate with helper springs on the shock or larger torsion bars
If you use helper springs it will raise the rear a lot: reindex the torsion bars or try some shorter, stiffer springs to see if you can get the ride height you want.
Also to calculate the effective spring rate add the torsion bar and the helper spring rate
e.g. 25.5 torsion bar and 100lb helper:
175 + (100 * 0.65) = 240lb effective rate
10-13-2009, 11:59 AM
#20
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the website and Paragon are incorrect?
what is that factor? the moment of inertia? the centroid? friction/shock compensation?
so their 240 becomes
e.g. 25.5 torsion bar and 100lb helper:
175 + (100 * 0.42) = 217lb effective rate
significant difference
what is that factor? the moment of inertia? the centroid? friction/shock compensation?
so their 240 becomes
e.g. 25.5 torsion bar and 100lb helper:
175 + (100 * 0.42) = 217lb effective rate
significant difference
10-13-2009, 03:26 PM
#21
The motion ratio of the spring is the distance from the pivot point of the lower control arm to the center of the spring divided by the distance from the pivot point of the lower control arm to the center of the tire.
wheel rate = (spring rate) * (motion ratio)^2
Why is motion rate squared? Because it's a force, and the lever arm is multiplied twice. Look below and see F1 and F2 in the equations, both have the motion rate in them so you wind up with the motion rate squared.
F1 one is the effective force of the spring using the ratio of the distance (motion ratio).
F1 = k1(L1/L2)dw
F2 using the F1 force to determine the "moment balance" using the ratio of the distance (motion ratio)
F2 = (L1/L2)F1
Book on the subject: "Racecar Vehicle Dynamics" by Milliken.
Lots of people have measured the motion rate on their 944 cars and as expected, there is a range of numbers. The range I have seen are:
Front 0.97- 0.91
Rear 0.61 to 0.65
From the link in post #14:
So how does all this relate to spring rates? Spring rates are usually expressed as lbs/in. In other words, if we compress a given spring by one inch, how much resistance does it offer? If a spring that is compressed 1" pushes back with a force of 500 lbs than that spring is rated as 500 lbs/in (assuming it is linear).
What we want to do here is determine a similar rate for the rear wheel. How much force does it take to push the rear wheel up a distance of one inch? The force required will be equal to the "effective rear wheel rate". A torque balance on the trailing arm will help us answer this question.
If we push up on the trailing arm and displace it by 1 inch (dw = 1" in Figure 1) - how far does this compress the spring (ds in Figure 1)? From similar triangles we calculate ds as:
L1/(ds) = L2/(dw) which leads to: ds = (L1/L2)dw
We will call the spring rate of the rear spring k1 (lbs/in). The force exerted on the trailing arm by the compressed spring is:
F1 = k1(ds) = k1(L1/L2)dw
Now use the moment balance on the trailing arm to calculate F2:
F2 = (L1/L2)F1 = k1(L1/L2)²dw
Divide through by dw to yield the "effective rear wheel rate":
Wheel Rate = F2/dw = k1(L1/L2)²
K1=Spring rate
L1/L2 =Motion ratio
So Wheel rate = Spring rate * (motion ratio)^2
wheel rate = (spring rate) * (motion ratio)^2
Why is motion rate squared? Because it's a force, and the lever arm is multiplied twice. Look below and see F1 and F2 in the equations, both have the motion rate in them so you wind up with the motion rate squared.
F1 one is the effective force of the spring using the ratio of the distance (motion ratio).
F1 = k1(L1/L2)dw
F2 using the F1 force to determine the "moment balance" using the ratio of the distance (motion ratio)
F2 = (L1/L2)F1
Book on the subject: "Racecar Vehicle Dynamics" by Milliken.
Lots of people have measured the motion rate on their 944 cars and as expected, there is a range of numbers. The range I have seen are:
Front 0.97- 0.91
Rear 0.61 to 0.65
From the link in post #14:
So how does all this relate to spring rates? Spring rates are usually expressed as lbs/in. In other words, if we compress a given spring by one inch, how much resistance does it offer? If a spring that is compressed 1" pushes back with a force of 500 lbs than that spring is rated as 500 lbs/in (assuming it is linear).
What we want to do here is determine a similar rate for the rear wheel. How much force does it take to push the rear wheel up a distance of one inch? The force required will be equal to the "effective rear wheel rate". A torque balance on the trailing arm will help us answer this question.
If we push up on the trailing arm and displace it by 1 inch (dw = 1" in Figure 1) - how far does this compress the spring (ds in Figure 1)? From similar triangles we calculate ds as:
L1/(ds) = L2/(dw) which leads to: ds = (L1/L2)dw
We will call the spring rate of the rear spring k1 (lbs/in). The force exerted on the trailing arm by the compressed spring is:
F1 = k1(ds) = k1(L1/L2)dw
Now use the moment balance on the trailing arm to calculate F2:
F2 = (L1/L2)F1 = k1(L1/L2)²dw
Divide through by dw to yield the "effective rear wheel rate":
Wheel Rate = F2/dw = k1(L1/L2)²
K1=Spring rate
L1/L2 =Motion ratio
So Wheel rate = Spring rate * (motion ratio)^2
Last edited by Bri Bro; 10-13-2009 at 10:38 PM. Reason: Added information
10-13-2009, 11:01 PM
#24
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went thru the derivation in post 14...
F1 L1 - F2 L2 = 0 or F1 = L2/L1 F2, a simple shear diagram
I can see where he factored in the incremental movement...to avoid the calculus to derive the moment (or force) diagram...
if I look at it like a pinned joint or cantilever, the distances are squared...because you integrate over the length...
his derivation is confusing but seems correct to me...
but as he stated, others take exception...
here's a basic derivation of a cantilever, without the ratios (or double cantilevers net moment)
x = distance from the tip
Wa = the uniform load at the cantilever end
V = Shear force
M = bending moment
1. V = -(Wa)x
2. M = SUM Vdx = SUM(-Wa)xdx = (-Wa)(x^2)/2 + C
Where; C = constant of integration; SUM means integral of
When x = 0; M = 0; C = 0; Hence;
3. M = -Wa(x^2)/2
Equation 3 is the equation of the bending moment at the cantilever portion
much clearer, at least to me...it's same if the ratio of the 2 levers are used
F1 L1 - F2 L2 = 0 or F1 = L2/L1 F2, a simple shear diagram
I can see where he factored in the incremental movement...to avoid the calculus to derive the moment (or force) diagram...
if I look at it like a pinned joint or cantilever, the distances are squared...because you integrate over the length...
his derivation is confusing but seems correct to me...
but as he stated, others take exception...
here's a basic derivation of a cantilever, without the ratios (or double cantilevers net moment)
x = distance from the tip
Wa = the uniform load at the cantilever end
V = Shear force
M = bending moment
1. V = -(Wa)x
2. M = SUM Vdx = SUM(-Wa)xdx = (-Wa)(x^2)/2 + C
Where; C = constant of integration; SUM means integral of
When x = 0; M = 0; C = 0; Hence;
3. M = -Wa(x^2)/2
Equation 3 is the equation of the bending moment at the cantilever portion
much clearer, at least to me...it's same if the ratio of the 2 levers are used
10-14-2009, 09:01 AM
#25
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Thanks Brian,for the illumination.
i have taken the factors so far for granted,as i believe did a lot of us on this forum.
the correct REAR factor is 0.42 and not any of the others.
Very interesting indeed.
i have taken the factors so far for granted,as i believe did a lot of us on this forum.
the correct REAR factor is 0.42 and not any of the others.
Very interesting indeed.
12-25-2009, 03:01 PM
#26
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Brian, you quote front motion ratios of .97 to .91. I assume that to get the effective front spring rate you would square the front motion ratio? So that using .94 as the front motion ratio (average of the quoted measurements), we'd have a front factor of .88.
12-25-2009, 11:10 PM
#27
Yup, you need to square the motion ratio to get the spring rate. There is the angle of the spring to control arm in the equation. Most of the time it is left out since the cosine of a small angel is close to 1 and has little effect on the result. I found this with Google, it has some nice drawing of the Angle correction.
http://www.gorancho.com/docs/spring_selection_guide.pdf
If the spring angle is 3 degrees from the control arm, the cos(3)= 0.99863. And that is damn close to 1.
http://www.gorancho.com/docs/spring_selection_guide.pdf
If the spring angle is 3 degrees from the control arm, the cos(3)= 0.99863. And that is damn close to 1.
Last edited by Bri Bro; 12-26-2009 at 12:33 AM.
12-26-2009, 12:21 PM
#29
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more seriously, if you are dealing in street rates, there is probably not much difference caused by leaving out the square of the motion ratio up front. But when you start dealing with a track setup where you might be talking about 750 or more front spring rate, a 10% reduction in effectiveness begins to yield a substantial amount of lbs/in.
02-17-2010, 01:45 AM
#30
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Hold up
Base S2 - 24mm torsion bar effective rate 139 lb/in
Base S2 - frt spring 138 lb/in
If this is true, isn't this a 1:1 frt/rear. For the stock car. Is the target ratio frt/rear 0.6-0.7 to reduce understeer (go more neutral)?
Or are my above numbers incorrect?
Base S2 - 24mm torsion bar effective rate 139 lb/in
Base S2 - frt spring 138 lb/in
If this is true, isn't this a 1:1 frt/rear. For the stock car. Is the target ratio frt/rear 0.6-0.7 to reduce understeer (go more neutral)?
Or are my above numbers incorrect?