ok I need to figure out springs for a formula student car.

we have a wheel weight of 170# with a pushrod at a 40* angle from horizontal. the rocker arm is a 2:1 (2" pushrod movement for 1" or shock movement)

so....

170#/(sin(40)) = 264 <------weight at rocker

264#*2=528# <------------weight at shock

is this right?
so a spring that has no preload would need to be huge then? sine the shock only moves .5" when unloaded then we need a spring in the 1000# range?

this does not seem right at all....


Nick

You have a total of four leverages being applied between the wheel and the spring:

1. The rocker arm (as you mentioned).

2. Pushrod angle (ditto, but usually expressed as a function of the angle from vertical; let's use the cosine to keep it simple but authors vary in their tables).

3. The ratio of the length of the lower wishbone to the distance of the pushrod mounting from the chassis pivot.

4. The ratio of the swing axle length to the length of the instant center of the lower ball joint.


Most of these leverages change in dynamic movement. Frankly, it's too friggin' complicated for me. So, I recommend drawing it up (maybe using the static numbers as a starting point for a rocker arm ratio) and determining your actual motion ratio (wheel movement divided by shock movement). This figure can be squared and applied to your desired wheel rate to get a spring rate.

Example (with some guesstimates of course):

L1 = 2
L2 = 1/cos(50) = 1.556
L3 = 18/16 = 1.225
L4 = 84/80 = 1.05

Total leverage = L1 * L2 * L3 * L4 = 4.00

The needed spring rate will be your desired wheel rate times leverage squared, or 16.

There's substantial support for basing wheel rate on a frequency target, or you can work out how much you want your wheel to travel in response to a shock. I'm going to use a wheel frequency of 120 cycles/minute, typical for a small racer with no ground effects (gleaned from Staniforth).

WH = wheel rate
SW = sprung weight = 170#

187.8 * SQRT(WR/SW) = 120
WR = (120/187.8)^2 * 170 = 69.4

Your needed spring rate would then be 69.4 * 16 = 1110

I think it's pretty funny that all this crapola produced pretty much the same result as your estimate. Did anbody say, "Just build it!"?

Things to look at are the rocker ratio and the pushrod angle, which seems very shallow.

Again, this is a preliminary estimate, and you should really check your motion ratios by drawing your suspension and testing it in one-inch intervals (to see if the suspension rate rises, and how much).

Hope this helps!

Pete

[Edited on 1/5/04 by pbura]

quote:

---

Originally posted by Syd Bridge
Still overcomplicating things Pete?

---

Hmmn

If correct a 1000lb spring rate will cause you no end of grief, ie you'll never find a small damper to effectively damp such a powerful spring.

Rather than calculating spring rate around your ride height, you should be looking at what you want in travel. Ride height can be fiddled with pushrod length and/or spring preload (as long as you keep a close eye on the geometry changes that can result).

FWIW, the FSAE team I've been helping out with this year have had custom coils wound between 60 and 100 lbs/in (with substantially different arm and rocker geometry).

And, just to be difficult, why are you doing this now? The competition is in less than 3 weeks!

If you've already got shocks and springs, I would suggest modifying your rocker ratio to suit -- a bit of a compromise, but the clock's ticking.

As a start, you could try flipping your existing rockers from 2:1 to 1:2 --- worth a try as you already have the bits.

Good Luck, see you in Pontiac, Ted

[Edited on 1/5/04 by andkilde]

yeah it is kinda late but...
the guy incharge of the application (I am new to team) filed it late (the USA event filled in less than one week) and we applied in october. so we are on the ticket for the UK race but are lacking the funds and support from the uni to ship the car over. and most of the car is not complete so it is not even competing this year at all. the designer of the car didn't design when he said he was....left the frame 75% finished and only had geometery for suspension (rockers floating in mid air in CAD) so in teh last few weeks we have been scrambling to finish the car and all. then we ran into redtape at school we had to fight for a month for our engine to get worked over. so we haven't shipped the motor for mods yet and that is gonna take 4 weeks on its own anyway.

well we are running some nice penske remote nitrogen shocks.

I could be wrong ont eh rocker arm and if it is 1:1 that should bring down the rates alot more.

I don't have the suspension geometery to look at effective swingarm length and all but being the arms are about 15" long and both mount inside the wheel so the ratio of the two woul dbe very near 1 anyways.

btw I do have the staniforth book in my own collection and knowing a better cycle rate is a big thanks (I was going to use 130 cpm)

Nick

[Edited on 5/1/04 by merkurman]

Nick,

The 120cpm for the front was from the Gould Terrapin, weight distribution 384F/676R, and it used 130cpm in the back.

Good luck!

Pete

well I'll look at what we got monday (car not all finished yet) and see if we can change to a 1.1 setup and get that rate down. I wish we had a more complete car so we could bolt up what we got and see how that is (we have a 150,200,250, and 350 spring)

would goign to a large preload help things at all or would the rates just be far too soft? IE bottoms out like crazy?

Nick

I thought that you might be using a bike shock with a 500+ rate, where the suspension leverage would be so great that you would be unlikely to run out of shock travel, but this may not be the case, judging by these spring rates.

Dampers are designed with a spring rate, or range of rates, in mind so that damping will be effective. If you use a shock made for a 100# spring with a 500# spring, you will be way under-damped. I'm going to assume that the springs came with the shock, and that damping is sufficiently adjustable in compression and rebound to deal with any one of these spring rates. In that case, you want to select a spring rate that will use as much of the stroke of the damper as possible, for most efficient damping, seeing as you have a choice.

Let's say that you want to provide for 5" of wheel travel, and that your shock travel is 3". In that case, you would want a motion ratio of 5/3, or 1.667. The resulting ratio of spring rate to wheel rate would be 1.667^2, or 2.778. As a result, the ideal spring rate would be 69.4 * 2.778, or 192#. Using this information, you would design around the 200# spring, and the corrected ratio of spring rate to wheel rate is 200/69.4 = 2.882. This figure equals the square of your planned motion ratio, making the ratio, or total suspension leverage, 1.698.

You've said that the only leverages of consequence are the pushrod angle and the rocker arm ratio, so if we divide needed total leverage by the pushrod factor, we'll have the rocker arm ratio. So, 1.698/(1/cos(50)) = 1.09, the theoretical ideal rocker arm ratio.

One advantage of the rocker arm system is that you can make your suspension adjustable. I suggest making your rocker arm (if the 3" shock stroke is correct) with three sets of holes, for 1.0:1, 1.1:1, and 1.2:1 ratios, for example. You could also swap in the 250# spring if you need. You could not use the 150# spring, as you would definitely run out of suspension travel.

We haven't talked about the effect of the angles of your rocker arm legs to the pushrod and to the shock, or their length, but these can add more leverage to your suspenson and give you increasing or decreasing rate. The examples given so far assume 100% efficiency of the rocker, or that the legs be at right angles to the pushrod and shock at all times, which is not possible.

If you start out with the leg angles at 90 degrees, you will have decreasing rate, with the amount of decrease governed by the radius of the arcs described by the mounting holes for the shock and pushrod. Another can of worms.

To keep it simple: Make the legs as long as possible, and arranged so that they will be perpendicular to pushrod and shock at full bump. That way, you can only have increasing rate. Another option would be to have one leg start at perpendicular, and have the other leg approach perpendicular at full bump. This would give roughly a constant rate.

Your leverages, and rocker arm ratios, will have to be adjusted for the differences in angle each leg is from perpendicular at rest.

About shock pre-load: Preload doesn't affect spring rate in any way, EXCEPT when you are using variable-rate springs that become stiffer as they are compressed. It's usually just a way of compensating for the unsprung weight of the car, so you don't use up shock travel just sitting there. At rest, about 2/3 of your shock piston should be showing.


Pete

Every race car build book has slightly different formulas, baised on the suspension geometry, for calculating spring rates. etc.
However, the easy way, and the most accutrate is to measure the shock/coil change in length for a given vertical movement of the wheel. This method will take into account everything including shock angle, etc.
Wheel movement/spring movement is the mechanical advantage ratio or suspension leverage (SL). [so if 1" wheel = 1/2" spring, SL = 2 and SL squared = 4)
The forementioned books show that a race car without significant aerodynamic loads (no wing or under chassis tunnel) should have a wheel frequency in the range of 100-125 CPM. As I got the front spring rate way off at 200#/in, back when I started, but got the rear rate of 150# about right (as determined by standing and bouncing on the tub for a rough feel! :-)
Anyway, I checked the rear WF (wheel freq) by determining the as raced weight, front and rear divided by 2 for each corner - then minus the sprung wieght of the tire, wheel, hub, and axil + shock/spring half weight (34# F and 39# rear in my case). This gave me a corner sprung weight of 142# F and 236# R, for a 900# as raced weight (wet w/200# suited-up driver and 1-1/2 gal of gas) with a 39/61 weight distribution.
Now for the formulas (Staniforth):
WF = 187.8 X SQUARE ROOT OF[ WR/SW]
where WF is wheel freq in CPM; WR is wheel rate in
#/in; and SW is sprung weight in lbs.

WR = wheel rate (#/in)
CR (coil rate) = WR X (SL squared)

For my 150#/in rear spring, this meant the rear WF is 127 CPM (on the high side of the acceptable range)
I then knew that the front should be slighty (5-10%) lower or say 120 CPM.
Working the formulas backwards, I get 127#/in or rounded off to 130 #/in for the needed fronts!!!!

John McCann

[Edited on 24/3/06 by jmccann]