Rear coil spring location vs rate revisited

[Egz]

Its been a while since I've had high school physics, so I want to make sure I got this right.

I want an effective spring rate of 400# (was 350#, but I think I want more) at the OEM rear spring location. However, I want to move it outward to the shock location. I have two answers for the correct spring rate, 200# and 300#.

My references are these two quotes from here

Originally Posted by covert1

Maybe this diagram will serve as a start: This happens to work out nicely for this discussion. This diagram is a balanced rod - in the sense that the forces upward and downward cancel each other out. Relating this to STOCK versus Relocated Spring location... If there's a 300# spring at the 8 unit position (STOCK) - the torque value is 2400 #-units. To move the spring (by applying opposite force at the 16 unit position (SHOCK Position) one must supply an excess of 150# of force. 150# spring at a distance of 16 units from fulcrum, you get 2400 #-units of force. Therefore, as OMNI said - if you want things being equal, put a 150lb spring in the shock location.

Originally Posted by P-51

If the distance between the inner pivot on the Focus arm, to the center of the spring perch is equal to 1, the distance from inner pivot to outer pivot is ~1.4.

I have the Ford drawings.

The wheel rate is equal to the motion ratio SQUARED, as has been said a couple times here. Thus, 1.4^2=2. If you're happy with an inboard mounted 350lb/in spring, you need to replace it with a 175 lb/in spring.

Period.

My rough measurements was the center of the OEM location from the pivot is 13.5", and the shock from the pivot is 18". 18/13.5 is really close to P-51's 1.4 number. But he said that value is squared. Now, if I use that 1/2 multiplier, I get 200#. But if I follow covert's picture, I get 300#. ( (400*13.5)/18 = 300 )

Any idea of which spring rate I should be looking at?

Thanks

[Stolle]

lost after 30 sec. lol

[Z63R]

Egz

Any idea of which spring rate I should be looking at?

P-51's because covert's drawing is oversimplified for the sake of discussion...

covert1

If there's a 300# spring at the 8 unit position (STOCK) - the torque value is 2400 #-units. To move the spring (by applying opposite force at the 16 unit position

...whereas P-51's stated dimensions more accurately represent the Focus's lca.

P-51

If the distance between the inner pivot on the Focus arm, to the center of the spring perch is equal to 1, the distance from inner pivot to outer pivot is ~1.4.

[Egz]

I'm just having a tough time finding out why he says the wheel rate is the motion ratio squared. I've gone over this several times, and all I can get is linear.

Edit: I found a page saying the same thing, and with a nice picture: http://www.miracerros.com/mustang/t_wheel_rate.htm

I've also seen several other pages saying the same thing. I guess its the scientist in me that is curious how that equation is derived. So I guess I'll aim for a 200-225 spring rate. I want to make a couple more measurements first, and see how much of an angle the springs are, and their effect on the rate.

[OmniFocus]

I can't for the life of me see how it can't be linear, but I'm thinking about it...

[Reza]

Ask the suspension god




The same applies to swaybars also the effective rate for the swaybar is (0.5)^2 is about 0.25*Swaybar Rate.

[Z63R]

The same applies to swaybars also the effective rate for the swaybar is (0.5)^2 is about 0.25*Swaybar Rate.

In plain English for we laymen, please.

[Johnny_P]

you changed your axis, WRC...

but it IS close enough to forget it. The struts are mostly 90*, and it doesn't make a difference if you're off by a small amount due to small angle sin approximation.

For anyone wondering, here's the equation with damper angle added in. Use WRC's pictures for reference on variables:


Looks like you want a 225 lb/in spring.

[Reza]

Originally Posted by Johnny_P

you changed your axis, WRC...

but it IS close enough to forget it. The struts are mostly 90*, and it doesn't make a difference if you're off by a small amount due to small angle sin approximation.

For anyone wondering, here's the equation with damper angle added in. Use WRC's pictures for reference on variables:


Looks like you want a 225 lb/in spring.

where did you get that formula?

[Johnny_P]

oy I just threw out my note sheet, lol.... lemme look for it

[Johnny_P]

I wrote the eq in MS Equation 3.0 for Word, lol
let me know if I did something wrong tho

[Reza]

you are right about dispalcement I over simplfied it.

But Its.

1. Cos (90-alph) = X2/A is not right the triangle doesn't have a 90 degree.


2. (K2) (X2) (D2)x (1/sin (alph) ) = (K1)(X1)(D1)

not the way you have it.

[Johnny_P]

I see where we're different. Your free body diagram is different than mine.

but,
sin(a)=opposite/hypotenuse
sin(a)=Vertical/F2
Vertical = F2*sin(a)

[Johnny_P]

regardless, you can omit that term because it will drop out, sin(90)=1. So, you're left with

k2= k1(D1/D2)^2

[Reza]

[F2 / sin(a)] = vertical component of F2

Not the other way around

[F2/ sin(a)] X D2 = F1 X D1

[Johnny_P]

No dude. Look at my drawing, then look up the definition of sin if you don't believe it. We may differ here because we drew our free body diagrams slightly different. But I assure you that:

sin(angle)=Opposite side of the triangle / hypotenuse of the triangle

In this case,
angle = alpha
opposite side = vertical side
hypotenuse of the triangle = F2

http://www.mathwords.com/s/sohcahtoa.htm

[grbrgrnzx5]

Anyone in here remember SOHCAHTOA?

Sine=opposite/hypotenuse

Just like Johnny P said.

[Johnny_P]

Its a moot argument anyway though. The dampers in the rear are *mostly* straight up, or 90*. sin(90) = 1, and cos(90-90)=1. So that whole last term drops out anyways.

To answer the original question, the answer is 225# springs. Although my equation differs from WRC's, they both produce the same answer either way b/c of that term dropping out.