11 replies to this topic

[meb58]

A Mac Strut is near universally described as having a 1:1 motion ratio. Wouldn't the strut need to be mounted in line with the tire's center line to truely be described as having a 1:1 motion ratio? I assume the 1:1 is derived from a minscule number that is rounded up, for the lack of a better phrase.

But is this number really so small as to be meaningless when increasing track width or changing camber or both? It would seem to me that at some point the numbers might grab one's attention...

Lets say I want to increase total track width by 300mm for goofs and giggles. And here is where I get into trouble...I see formula for calculating Mac Strut motion ratio but these seem to be concerned with the outer ball joint location and not the tire's center line. This just seems wrong to me since the tire's distance from the inner control arm pivot point changes its leverage on the strut location at the LCA.

[gruntguru]

A Mac Strut is near universally described as having a 1:1 motion ratio. Wouldn't the strut need to be mounted in line with the tire's center line to truely be described as having a 1:1 motion ratio? I assume the 1:1 is derived from a minscule number that is rounded up, for the lack of a better phrase.

But is this number really so small as to be meaningless when increasing track width or changing camber or both? It would seem to me that at some point the numbers might grab one's attention...

Lets say I want to increase total track width by 300mm for goofs and giggles. And here is where I get into trouble...I see formula for calculating Mac Strut motion ratio but these seem to be concerned with the outer ball joint location and not the tire's center line. This just seems wrong to me since the tire's distance from the inner control arm pivot point changes its leverage on the strut location at the LCA.

Unless there is significant camber change during travel, the tyre's centreline is irrelevant because the wheel translates in parallel with the ball joint regardless of how far it is from the ball joint.

Edited by gruntguru, 19 June 2009 - 04:38.

[Greg Locock]

It pretty much depends on the angle of inclination of the strut. When in doubt draw the thing out in front view in 2 positions and work out the change in length of the strut compared with the change in height of the wheel. That is all it is.

[cheapracer]

To add to Greg there is an arc travel and the wheel usually goes out and up firstly which will give the wheel more vertical travel than occurs at the shaft.

You simply couldn't do better and faster than grab 2 pieces of 1 x 1 wood and use some nails, 1 pivot in the ground for your LCA, 1 to pivot the LBJ (join the 2 pieces of wood) and 2 at the top as guides - this will enable you to do some quick measuring, measure the 'wheel travel' against how far the wood draws thru the nail guides.

What is good about it all even though the ratio isn't far off 1:1 is that it usually is an increasing rate ie; starts out soft and progresses to harder (not a big change though).

Edited by cheapracer, 19 June 2009 - 09:58.

[meb58]

Okay, I understand how to measure this now. The arc cheapracer is describing is from the instant center, yes?

And pardon my thick headedness, I have to ask again and this regards gruntguru's reply. If I construct a twice life scale of the strut and flip it up side down, like a one seat see saw and press down on what is the tire's center line, then move the tire's center line 50% farther from the inner pivot, why wouldn't the force required to push down on the strut be half?

If we flip the strut back over I see the ground pushing up on the tire...

Or am I mis-applying leverag equations? This is rudimentary, i know...I refer you back to my thick headed note...

[Greg Locock]

A macpherson is a funny linkage, the strut is directly fixed to the wheel, and for most applications the strut is more or less vertical, hence MR=1. As you lean the strut over more this relationship gets upset, but you rapidly degrade the smoothness of the mechanism (for instance if you lean the strut right over parallel to the lower arm the thing becomes a variable length swing arm - no use to man or beast.

[gruntguru]

I have to ask again and this regards gruntguru's reply. If I construct a twice life scale of the strut and flip it up side down, like a one seat see saw and press down on what is the tire's center line, then move the tire's center line 50% farther from the inner pivot, why wouldn't the force required to push down on the strut be half?

If we flip the strut back over I see the ground pushing up on the tire...

Or am I mis-applying leverag equations? This is rudimentary, i know...I refer you back to my thick headed note...

If the tyre (wheel) was fixed to the control arm (ie swing axle type suspension) you would be correct - double the distance from the pivot to the wheel -> double the wheel travel -> half the wheel rate. But the wheel is fixed to the strut. If there is no camber change, the wheel travel is the same no matter how far you offset it from the car.

Imagine you have dual wheels with a tyre only fitted to one of them. If they "translate" up and down together (no camber change), it doesn't matter which wheel you fit the tyre to, the travel for a given spring displacement and therefore the "wheel rate" is the same for both. In the case of the "swing axle suspension" (lots of camber change) the wheel travel for the two wheels is very different and so the wheel rate will also be very different.

[Powersteer]

for instance if you lean the strut right over parallel to the lower arm the thing becomes a variable length swing arm - no use to man or beast.

So if the chassis flex the suspension will be all over the place. So harder bushings and strut cross bracing would help a lot then.

[cheapracer]

Okay, I understand how to measure this now. The arc cheapracer is describing is from the instant center, yes?

The arc is controlled by the LCA pivot, from there the length of the LCA to the LBJ, the LCA's starting angle, the pivot or determining guide at the top of the strut, the distance that is from the LBJ as well as the distance from the wheel (vertical) centerline and thats only 2 dimensional speak - you must add a 3rd dimension (or not) depending on where your 2nd LCA pivot is or track rod is mounted.

LCA = lower control arm
LBJ = lower ball joint

[Greg Locock]

Here's a rule of thumb, for a fairly conventional looking geometry, MR=0.93, move the top of the strut in by 100 mm, MR=0.83

I realise that's not quite what you were asking.

If on the other hand you move the wheel out by 150, the MR of the original strut geometry changes to 78% (that is effectively applying an enormous offset to the wheel, only).

[Greg Locock]

Neil reminds me that the demo version of susprog3d includes a macp setup, so i suspect you could answer many of these questions with that.

[meb58]

I guess I was using Swing arm logic

Greg, I wasn't asking exactly that, but your reply was very helpful...huge, perhaps unrealistic changes do not upset Mac Strut MR all that much. So 5-15mm is nearly insignificant...even when combined with a little more neg camber.

cheapracer, gruntguru, thank you...the dual wheel example is a good one.