So you thought you would never use high school geometry!! I am trying to write an excel program to calculate the instant center of the associated dynamic front end for both static and dynamic conditions. My problem is how do you model the fixed lower arm to determine the instant center? For hinged upper and lower arms I extend the lines of the arms to find the intersection of the two but what do you do with a fixed lower arm. I tried just extending a horizontal line through the lower arm pivot point and it looks OK for the static calculations but doesn't look right for the dynamic calcs. Any suggestions?

I tried doing the same and my conclusion was the same. But if you really look closely at the situation, it does work. It is easy to lose sight of the actual condition of the car when the suspension is compressed. I was trying to view the system the same way as if it was at rest, when it actually is not. You need to compensate for the lower chassis height when the suspension is compressed. When the suspension is compressed, the car is lower to the ground. Create a a constant that takes this into acount and apply it to both situations. For the static measurements, it will have a net zero effect, but for the dynamic measurements it will make the numbers come out correct.
Hope that helps.
Rick<><

ps: I wrote a formula for calculating spring tension when you move their position relative to the pivot pin location (effectively lengthing or shortiening the upper arms), useful for prototyping new front ends. It also accounts for using more than one spring also. I'd be glad to share it with you.

ps: I wrote a formula for calculating spring tension when you move their position relative to the pivot pin location (effectively lengthing or shortiening the upper arms), useful for prototyping new front ends. It also accounts for using more than one spring also. I'd be glad to share it with you.

If you wouldnt mind sharing can I get a copy of it??? Ive messed around with upper arm legnth and made my own judgement calls as to what I was gaining through the change in roll center and camber gain..But as far as spring rate change I just used the general assumption that the longer the arm the softer the spring would act. It would be interesting to look at hard numbers to see if my changes were correct when legnthening the upper arms..

The longer the arm the softer the spring you'll need to use because it acts like its harder. Go to my website, and send me and email and by the afternoon I'll send you a spreadsheet. I was using Mathcad, But I can write it in excel, or just post it as aformula that you can use a calculator with.
Rick<><

I don't know if this is what you're looking for, but maybe it'll help.


Try this:

F1 = M(d1/d2)-1

M= Original spring rate

F1 = The new spring rate

D1 = The original distance from the pivot pin, (through the castor block) to the spring (or Kingpin)

D2 = The new distance from the pivot pin


Example: Current spring…. 15lbs
Current distance 1.125”
New Distance is 1.0 “

(1.0 /1.125)-1 x 15 = 16.875

The New Spring will have to be 16.875 lbs.

Keep in mind, the "-1" means to the power of negative 1
Hobbytalk doesn't allow subscripts and superscripts so that the formula actually looks like a formula.

Ive messed around with upper arm legnth and made my own judgement calls as to what I was gaining through the change in roll center and camber gain..But as far as spring rate change I just used the general assumption that the longer the arm the softer the spring would act. It would be interesting to look at hard numbers to see if my changes were correct when legnthening the upper arms..

There wouldn't be any change in spring rate by changing the length
of the upper arm. It does not figure into the equation, as our application
it is a locating device, not a true suspension member.
The only thing that would change the 'wheel rate' is spacing the wheel
out or in on the axle....

So you thought you would never use high school geometry!! I am trying to write an excel program to calculate the instant center of the associated dynamic front end for both static and dynamic conditions. My problem is how do you model the fixed lower arm to determine the instant center? For hinged upper and lower arms I extend the lines of the arms to find the intersection of the two but what do you do with a fixed lower arm. I tried just extending a horizontal line through the lower arm pivot point and it looks OK for the static calculations but doesn't look right for the dynamic calcs. Any suggestions?

I think your lower horizontal line just needs to extend thru the center of the lower pivot ball level with the chassis. Remember that the pivot ball moves with the chassis, not the wheel when you are trying to calculate dynamic centers

I don't know if this is what you're looking for, but maybe it'll help.


Try this:
That is an impressive spreadsheet. Who did this? and is there one available for a double a-arm suspension (off-road or dirt oval cars)?

There wouldn't be any change in spring rate by changing the length
of the upper arm. It does not figure into the equation, as our application
it is a locating device, not a true suspension member.
The only thing that would change the 'wheel rate' is spacing the wheel
out or in on the axle....

That's not entirely true. We all know the spring rate doesn't change (by changing the upper arm angle or length), we all use that term but really mean wheel rate. But the location of the upper arm inner pivot point does affect wheel rate. So if you have a long upper arm mounted low (towards the chassis), you will have a different wheel rate then if you have a short arm mounted high. Plus when the car rolls the effective force (angle) being applied to the wheel is different for the left vs the right front due to camber and chassis roll. IMO I'm not a Chassis Eng but I stayed at a Holiday Inn Express last night.

That is an impressive spreadsheet. Who did this? and is there one available for a double a-arm suspension (off-road or dirt oval cars)?

A guy named "Skull and Bones" did the spread sheet. If you do a search back on suspension geometry, there is a whole long thread he started on different setup stuff.

I don't know about a free program for full size cars other than Joel Browns, but I have the program Bob Bolles produces and also Performance Trends roll center calculator if thats what you're asking about. You can send your measurements and I'll be glad to run them through.

I should have specified, RC off-road or DO cars. Thanks for the info!!

Only problem with the one that Skull and Bones put out is the lower arm is not stationary, his model moves the lower as if it were able to pivot which does not give an accurate representation of where the roll center really is. On the upper a-arm thing, I would think the change in roll center location would have a much bigger effect on the "stiffness" of the front end as the spring rate wouldn't change at all just because you changed the length of the upper arm. On our big car we have run with the roll center close to the middle and up a little and the front end "feels" stiff, now move the roll center left 2 feet and down and then the front end feels soft with the same springs. When we changed roll center locations we had to run stiffer springs to keep the car up off the track but we didn't need as much rebound in the shocks to keep the car on the ground. That's my opinion anyway!!

I think people are getting confused between spring rate and wheel rate. The only thing that will change spring rate is pre-load (since our coil springs are not linear), or the spring itself. However, by changing the spacing of the tire on the axel, or by lengthening the upper arm, you do impose a change in wheel rate. If you lengthen the upper arm, the spring will utilize a longer moment on the upper arm, allowing it to exert more force (like a lever). However, by spacing the tire out, the tire imposes a larger moment on the upper arm. Both will not have the same affect on handling. (don't forget, the tire also acts as another spring)

The attached drawing should help clear up any issues on Instant Center and Roll Center.

One thing to remember guys, listen to your car before getting into all this geometry stuff. All this info can get you so lost that your car will never turn left again. If your cars say give me more right front spring, give it more spring, if it tells you it needs more right rear, give it more right rear.

I will agree knowing all this geo info does make you think about the car and how it works but don't get to caught up in this stuff, it will result in many sleepless nights, trust me, I have a few now and then thinking about this stuff.

If you are a little crazy and ready for some of those sleepless nights and you really want to try thing related to geo findings, one thing to remember, one geo change will likely effect a lot of things so you may have to make multiple changes to really feel the difference of a geo change. Also, be sure to re-check all your settings when making these changes, camber, caster, toe, ect. Simple re-checks like that can really make a difference.

Hope this helps guys.

That's not entirely true. We all know the spring rate doesn't change (by changing the upper arm angle or length), we all use that term but really mean wheel rate. But the location of the upper arm inner pivot point does affect wheel rate. So if you have a long upper arm mounted low (towards the chassis), you will have a different wheel rate then if you have a short arm mounted high. Plus when the car rolls the effective force (angle) being applied to the wheel is different for the left vs the right front due to camber and chassis roll. IMO I'm not a Chassis Eng but I stayed at a Holiday Inn Express last night.

Mike.. you are one of my r/c heroes... but I believe you are wrong on this. :rolleyes:
The upper a-arm on our cars, has absolutely no effect on the wheel rate.
Wheel rate, is what the car 'thinks' is in it for a spring.
We are all, I believe, on the same page with this..
It is calculated by finding the motion ratio, which is, the consideration of the
center of the tire contact patch, the length of the arm, and the location
of the spring, and taking this ratio and factoring it into the spring rate.
So for example only, say you have a motion ratio of .63,
then the wheel rate of a 12# spring, is say 5.5#
If you shim the wheel out, it may go to say 5.1#...
The change in the relationship, between the lower arm length,
and the distance between the kingpin (spring),
and the center of the tire contact patch,
are the only factors in changing the effective spring rate (wheel rate)...

All of the force input by the spring in bump,
is taken in by the lower, solid a-arm.
The upper a-arm, only keeps the kingpin where you want it.
It isn't really even necessary to have an upper a-arm,
to compute a wheel rate..
If you imagined CA gluing your lower pivot ball to the a-arm, so that
your kingpin was vertical, your wheel rate would be the same, regardless
of the absence of an upper arm...
If you shim the wheel out .060"... you absolutely change the wheel rate...

Let me ask you this, to illustrate the significance of the upper a-arm....
How would you calculate your wheel rate, with the above example?
How could the a-arm have an effect on it, if it were not even there?
It can't....

All of this talk about the length of the upper a-arm, and it's effect on camber
change, I agree with 100%. THAT is one function of the upper...
All of the talk about it affecting the roll center, I agree with, to a point.
But even just changing the length of the arm,
will not in itself necessarily change the roll center...
If you also change the inner mounting position,
you could end up with a different camber curve,
but the exact same roll center.. :thumbsup:

Thanks Guys. This has been enlightening. I got a copy of "Skull and Bones" spreadsheet but have been modifying it to fit my thoughts on the subject. If I'm ever satisfied with the results I'll post it here. Thanks again for the input.

I think people are getting confused between spring rate and wheel rate. The only thing that will change spring rate is pre-load ](since our coil springs are not linear), [/U] or the spring itself. However, by changing the spacing of the tire on the axel, or by lengthening the upper arm, you do impose a change in wheel rate. If you lengthen the upper arm, the spring will utilize a longer moment on the upper arm, allowing it to exert more force (like a lever). However, by spacing the tire out, the tire imposes a larger moment on the upper arm. Both will not have the same affect on handling. (don't forget, the tire also acts as another spring)

Just for clarity's sake coil springs are very linear and not linear, but in 2 different aspects. Within a ceratin range of displacement, a coil spring is very linear in that the relationship between displacement and force is very linear. What I am sure you're refferring to however, is that as you compress the spring, the force increases.
The formula for the spring force constant is K = FD

Dan I thought the same as you, until I was given a prototype front end. It alowed you to move the upper arm to almost any location.By doing so it also veried the wheel rate. With the standard ASC type front end I don't think the amount of movement lets us notice the wheel rate is changing when the inner upper arm pivot moves. Try this, put the upper arm pivot in the lowest hole and set the camber to zero degrees on the right front zero preload, put the left front in the upper hole on any block that rasies the pivot above the inner pivot with the same castor, camber, and preload settigns.You will notice it takes more force to push the left front tire into the spring as does the right front. As far as a geometric example I can't supply one, but know it happens. The Prototype front end never made it into production but was run allot on 1/12 scale road course cars. I was shocked by the result of moving the upper arm around and never really could explain why it happened, but I know it wasn't from any binding.

it does change wheel rate, simple test (which I have done). IRS caster blocks, use the all the way up hole, and outside hole (short arm), the hook a newton meter to the spoke of the wheel and pull on it, and see how much force it takes to compress the spring say 1/8". Then move the hinge pin to the inner hole (long arm) and re-adjust the camber. do the same test, and you will see a difference in the 2 numbers, the longer arm being the softer of the 2.

The shorter arm creates more dynamic camber. As you're pulling straight up on the axle with a higher rate of camber gain, you introduce more bind (resistance) at multiple points by increasing the king pin angle...which increasingly deviates from the force vector of the "string". Sorry, it must be getting close to carpet season.

Jason and Jeff, thanks for putting my info into English.

Jason and Jeff, thanks for putting my info into English.

now thats funny!! :hat:

The shorter arm creates more dynamic camber. As you're pulling straight up on the axle with a higher rate of camber gain, you introduce more bind (resistance) at multiple points by increasing the king pin angle...which increasingly deviates from the force vector of the "string". Sorry, it must be getting close to carpet season.

That paragraph has a 4 letter word in common with
the sketch I sent you Mike... :wave:

Thanks Guys. This has been enlightening. I got a copy of "Skull and Bones" spreadsheet but have been modifying it to fit my thoughts on the subject. If I'm ever satisfied with the results I'll post it here. Thanks again for the input.

I never could change the things that needed changed, how did you unlock it?