Calculating camber change with suspension travel
Discussion
If the software for a fully calculated solution isn't available, measuring the actual car would be the next simplest and most reliable approach.
If that isn't attractive, then you may still be able to get a rough idea using the 'back of a fag packet' approach I suggested. The more complex the system the more work will be required and the less accurate the answer will be, but it is possible to deal with things like wishbone axis not parallel to the centerline if you're willing to solve the problem in several stages.
Ultimately what you want is to know the sideways and vertical deflection of the top ball joint i.e. the path taken in a 2-d plane, and the same for the bottom ball joint. This gives you two arcs, and path taken by the upright is represented by pairs of points that are the appropriate distance apart. Once you have the two arcs calculated/plotted graphically, showing the path taken by the upright is a doddle. Or you can calculate it using trig, if you really want to do it the hard way.
To work out the arc followed by the end of a wishbone with a horizontal pivot axis not parallel to the centreline, you could start by pretending that it *is* parallel, and then scale the resulting lateral deflection by the sin of the actual angle (pivot axis to centerline). This is easy to do graphically (pencil and beermat) or by scaling the effective length of the wishbone up (90 deg pivot axis is equivalent to an infinitely long wishbone in terms of lateral deflection).
If that isn't attractive, then you may still be able to get a rough idea using the 'back of a fag packet' approach I suggested. The more complex the system the more work will be required and the less accurate the answer will be, but it is possible to deal with things like wishbone axis not parallel to the centerline if you're willing to solve the problem in several stages.
Ultimately what you want is to know the sideways and vertical deflection of the top ball joint i.e. the path taken in a 2-d plane, and the same for the bottom ball joint. This gives you two arcs, and path taken by the upright is represented by pairs of points that are the appropriate distance apart. Once you have the two arcs calculated/plotted graphically, showing the path taken by the upright is a doddle. Or you can calculate it using trig, if you really want to do it the hard way.
To work out the arc followed by the end of a wishbone with a horizontal pivot axis not parallel to the centreline, you could start by pretending that it *is* parallel, and then scale the resulting lateral deflection by the sin of the actual angle (pivot axis to centerline). This is easy to do graphically (pencil and beermat) or by scaling the effective length of the wishbone up (90 deg pivot axis is equivalent to an infinitely long wishbone in terms of lateral deflection).
Edited by GreenV8S on Wednesday 2nd September 00:07
Sam_68 said:
spend said:
Its not hard, there should be online theorums to calc intersection of the 2 prescribed circles.
Surely that only does for 2-dimensions? Fine if you've got no anti-dive/anti squat, and the pick-ups are parallel to the chassis centreline...Must admit, I'm happy with my computerised paper dollies, though - end results is all I'm interested in - I'll leave the hard stuff to you chaps!
ETA, this is the cludge I made to calc the bits for me:
- As a bit of explanation to this perly script:
- Ht is Ht of car you wish to eval, I change it increments bouncing between bump & droop to animate the model
- org_ht is Ht when lower wishbone is horizontal
- lwl is lower wishbone length
- dx is x offset of chassis upper wishbone pivot point
- dy is y offset of chassis upper wishbone pivot point
- upl is the length of the upright
- uwl is upper wishbone length
$ly = $Ht - $org_ht ; #change in ht of lower wishbone
$la = asin($ly / $lwl) ; # angle in rads
$lx = $lwl * cos($la) ; #
$d = sqrt( ($lx - $dx)**2 + ($ly - $dy)**2 ) ;
$a = ( $upl**2 - $uwl**2 + $d**2 ) / 2 / $d ;
$pia = acos( $a / $upl ) ; #int angle between upper pivot and upright
$uia = acos( ($lx - $dx) / $d ) ; #int angle horiz upchass - upright
$wua = $pia + $uia ;
$wux = $lx - ( $upl * cos($wua) ) ;
$wuy = $ly + ( $upl * sin($wua) ) ;
}
Edited by spend on Wednesday 2nd September 18:06
leorest said:
spend said:
Try:
http://www.racingaspirations.com/?p=286
I was going to make my own animation a while ago, but that site serves most purposes I think.
That is interesting. It's a shame it doesn't allow values to be typed in...http://www.racingaspirations.com/?p=286
I was going to make my own animation a while ago, but that site serves most purposes I think.
This is a DOS based program that allows you to enter the coordinates of the pivots. Works fine on Windows XP, not tried it on Vista.
Edited by Mr2Mike on Thursday 3rd September 10:23
I'm not sure if this helps you, but I had to finish what I was working on to help me design new wishbones - really wanting to use lighter uprights. I have an animated program that can be run on any system that has perl (&Tk module) installed - which are free 
It's just really a personal tool, so I just customize it when I want to look at specific things.
a couple of screen shots of the 2 main windows:
it will just calculate all scenarios at the incremental ride heights between full bump & droop, so appears animated, and displays some graphs of points of interest.
It's just really a personal tool, so I just customize it when I want to look at specific things.a couple of screen shots of the 2 main windows:
it will just calculate all scenarios at the incremental ride heights between full bump & droop, so appears animated, and displays some graphs of points of interest.
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