Rear ride height affecting camber? Dumbass I meant Caster D’oh!!

[72bluNblu]

Yes I agree, dividing the 1.7 by 108 does give that figure, which as you say equates to a tiny angle change. Thats the reason i ended up plotting it out in my CAD programme (I design exhibition stands), because when I put the ratio of V 1.7 (43mm) to H 108 (2743mm) into an online calculator it returned the degree value of 0.898.

View attachment 1716436429


So I drew up a very basic representation of the frame rails at horizontal and sitting at the current rear height, with a representative upper BJ set horizontally back from the contact point of the tyre to give a nominal caster of 3 degrees. Then using the centre of the front wheel as the pivot point I rotated the frame to reach the new height at the back +1.7" so the dashed horizontal line on the second drawing is the original line of the frame rail and the angled dashed line represents the frame angle with the rear lifted +1.7 the dimensioning tools in the CAD programme are generated by the programme not my maths so it automatically finds the angle between those 2 lines.

Also another thing I wasnt expecting, I dont know if you saw on the drawing, but the change in frame angle does not directly corelate to the reduction in caster, so the frame angle is +0.88deg but this only reduces caster by -0.29deg. I had assumed that it would be like tilting a table and the top and the legs would rotate by the same amount. In this case I believe that the reason it doesnt is that the caster angle is from the contact patch to the upper BJ whereas the frame angle is from the rear of the wheelbase to the centre of the front wheel so the origin points of those 2 angles are in different places creating 2 differntly proportioned triangles.

It was interesting Blu's comment above seems to tally in roughly with what I found from the drawing.


Anyway for all the drawings and calculators, the reality is I need to drop the rear height using the adjustable spring hanger because it looks silly and isnt as nice to drive as before. I doubt ESPO sent the incorrect springs their reputation tends to make me think not, but the springs that were on there must have been really worn out as the difference in height now is pretty big. To be honest going out and re checking from axle centre to arch with the wheel off its probably more like 2.5". Lucky I fitted adjustable hangers!

EDIT: SO I THINK I HAVE THE ANSWER!

SOH CAH TOA. sine would be for if i knew the opposite and hypoteneuse sides, I dont, I know the opposite and adjacent so I need tan, actually I need inverse tan. School was worth something.

View attachment 1716436430

When you lift the rear end, you are rotating the front of the car around the front axle, the spindles. Of course, that also happens to be the caster angle.

Maybe I am missing something, but I see the caster angle change to be the tan of angle 0.016, which also happens to be 0.016 degrees.

View attachment 1716436431

While I tend to be a bit of math nerd myself, I'm not sure that the calculations you guys are doing are representative. @Righty 's calculation of .3° for 1.5" seems to be in the ballpark for me, but I'm not sure his math/methodology is absolutely correct (it may be!).

But yeah, having actually MEASURED the caster change with a 2" change in the front/rear rake on my car and had the alignment come up with ~.5° of a difference, I would say the difference should be close to that ballpark with minor changes one way or another based on the particular set up of the car in question. It is a matter of tenths of a degree of change, which is significant. Not hundredths.

That would also be backed up by the noticeable change in steering characteristics, because losing a half degree of caster (or close to it) on a car that doesn't have that much to begin with would be noticeable.

I suppose I could put my car back on the turn plates and slip plates and throw a block on top of the slip plates and measure it again, but I don't think I'll have to the time to be doing that for a bit as I'll be working all weekend.

As Rusty Rat Rod often says a picture is worth a thousand words. Sorry it's raining here so the afters are garage pics. Here is a before on the left and after on the ride height with the ESPO +1" springs, as i said either they sent me +2" by mistake, (unlikely but possible) or my old springs were totally FUBAR. It's about 2.5" difference using the basic measurement of floor to arch lip. Shackles are very upright I loosened them and gave it a bounce but no change. I might loosen up the U bolts to see if it's bound up. On another note that touch up spray around the back corner of the door makes me dread what I'm going to find under there. Although the rails look pretty solid from underneath.

View attachment 1716436433View attachment 1716436434View attachment 1716436435

My guess would be that your original springs were completely worn out, and the +1" springs that ESPO sent you will probably settle another inch with some miles and time. All springs settle a bit.

So I just checked the suspension height at the front, from the bottom of the adjustment blade to ground is 8.7" and from the bottom of the LCA knuckle is also 8.7". According to the FSM the suspension height for the Barracuda should be 1 3/8" +- 1/8". So unless i am taking my measurements in completely the wrong place that would be 1 3/8 lower than factory spec interesting.

So that measurement is taken from the bottom of the adjusting blade (torsion bar adjusting lever) to the bottom of the lower ball joint housing. The Barracuda was a bit lower than other A-bodies, which run that A-B measurement at 1 7/8". Either way if you're getting 0 that means your car is sitting fairly low in the front, and it should sit higher if you're not running larger aftermarket torsion bars and have adjusted your bump stop heights.

Here's a blown up version of the FSM showing the measuring locations

But yeah, also not uncommon for these cars to be sitting low in the front. Especially if the stock torsion bars are still in place, torsion bars sag just like any other spring, they just do it radially. And that lowers the car.

 

[Bewy]

My comments & posts in this thread are based on the OPs statement that he was raising the rear by 1.7".........& he wanted to know what affect this would have on caster & camber.

 

[72bluNblu]

My comments & posts in this thread are based on the OPs statement that he was raising the rear by 1.7".........& he wanted to know what affect this would have on caster & camber.

Right, so my actual alignment measurements on a 2" raise in rear ride height being a ~.5° loss in caster are meaningless?

By all means, if you have actual measurements to back up your theory, let's see them. But I'm pretty certain your math isn't representative. @Righty 's drawings appear correct and the .3° change is certainly within the margins considering different ride heights and set ups.

Saying the difference is "unmeasurable" certainly flies in the face of everyone that's actually changed the ride height by that much and said that it resulted in noticeably different steering characteristics though. So let's see your evidence, not just your math.

 

[Bewy]

This is about geometry, already shown the maths, post #21. The rear of the car at the rear axle is being lifted 1.7", ie the body/frame is being lifted 1.7". And the body is pivoting around the front spindles; the suspension components that will be affected [ caster, camber ] are attached to the....body.
Lifting the body 1.7" over a length of 108" is a very small change, about 1/50 th in round numbers. So the caster change will be very small also....

 

[RustyRatRod]

Also remember, vehicle wheelbase will have an effect. The shorter the wheelbase, the more effect changing the rear height will have on caster.

 

[Righty]

Please excuse my ignorange!

From the posts of yours I've read on various topics I cant see why you'd apologise for ignorance!
My problem is I come on here as someone who truly has no knowledge on most of the things I'm asking about and then end up trying to grasp the concepts behind the answers, there must be over a millenium of combined knowledge on FABO, not that you always agree with eachother

 

[remytherat]

Raising the rear ride height will cause the chassis to pivot around the front spindle. If we pivot by 1.7" inches on a 108" wheelbase this draws out a triangle with side lengths 108 by 108 by 1.7 (the third side is slightly more than 1.7 because 1.7 is actually the altitude but this is a negligible difference). The opposite angle to the 1.7" side is 0.902˚, which is the amount the chassis pivots. Likewise, the line between the upper and lower balljoints will pivot by 0.902˚, which will be the resulting change in chaster.

 

[67Dart273]

Raising the rear ride height will cause the chassis to pivot around the front spindle. If we pivot by 1.7" inches on a 108" wheelbase this draws out a triangle with side lengths 108 by 108 by 1.7 (the third side is slightly less than 1.7 because 1.7 is actually the altitude but this is a negligible difference). The opposite angle to the 1.7" side is 0.902˚, which is the amount the chassis pivots. Likewise, the line between the upper and lower balljoints will pivot by 0.902˚, which will be the resulting change in chaster.

Do I understand? You are saying that if you lift the frame at the center of the rear axle by 1.7" it changes caster by "almost" 1 degree?

 

[Bewy]

I see that I have made a mistake in my calculations in my posts [ above ]. I get the same as Remy, 0.9 degree change in caster.

 

[Righty]

While I tend to be a bit of math nerd myself, I'm not sure that the calculations you guys are doing are representative. @Righty 's calculation of .3° for 1.5" seems to be in the ballpark for me, but I'm not sure his math/methodology is absolutely correct (it may be!).

But yeah, having actually MEASURED the caster change with a 2" change in the front/rear rake on my car and had the alignment come up with ~.5° of a difference, I would say the difference should be close to that ballpark with minor changes one way or another based on the particular set up of the car in question. It is a matter of tenths of a degree of change, which is significant. Not hundredths.

That would also be backed up by the noticeable change in steering characteristics, because losing a half degree of caster (or close to it) on a car that doesn't have that much to begin with would be noticeable.

I suppose I could put my car back on the turn plates and slip plates and throw a block on top of the slip plates and measure it again, but I don't think I'll have to the time to be doing that for a bit as I'll be working all weekend.


My guess would be that your original springs were completely worn out, and the +1" springs that ESPO sent you will probably settle another inch with some miles and time. All springs settle a bit.


So that measurement is taken from the bottom of the adjusting blade (torsion bar adjusting lever) to the bottom of the lower ball joint housing. The Barracuda was a bit lower than other A-bodies, which run that A-B measurement at 1 7/8". Either way if you're getting 0 that means your car is sitting fairly low in the front, and it should sit higher if you're not running larger aftermarket torsion bars and have adjusted your bump stop heights.

Here's a blown up version of the FSM showing the measuring locations
View attachment 1716436787


But yeah, also not uncommon for these cars to be sitting low in the front. Especially if the stock torsion bars are still in place, torsion bars sag just like any other spring, they just do it radially. And that lowers the car.

It’s a British import RHD Barracuda, these came with the Formula S brake and suspension package, but not the engine, so that should be the heavier duty torsion bars. Just catching on the posts above, I’d say I don’t have any confidence in my maths, but as a CAD Monkey of 30 years I’m pretty confident to back the results of what I drew up. I think the reason the angle of caster change is different to the angle of frame change is to remember that the two right angle triangles have their small acute angles originating in different places, the frame angle triangle originates at the front wheel hub centre, whereas caster angle originates at the contact patch of the tyre, effectively 12” vertically downwards from the hub centre. So as the car tips forward that tips the upper ball joint forward but not from the same pivot point. It’s really fascinating to see how some fairly small differences make quite noticeable differences.

 

[Righty]

My guess would be that your original springs were completely worn out, and the +1" springs that ESPO sent you will probably settle another inch with some miles and time. All springs settle a bit

I’m hoping so!

 

[Righty]

Also remember, vehicle wheelbase will have an effect. The shorter the wheelbase, the more effect changing the rear height will have on caster.

Yep shorter long sides to the triangle, I guess also weight transfer important , how much time do all the racing categories spend chasing consistent weight distribution, just visually it’s clear looking at my Barracuda that the increased rake must be tipping the weight distribution forward especially under braking.

 

[Righty]

Do I understand? You are saying that if you lift the frame at the center of the rear axle by 1.7" it changes caster by "almost" 1 degree?

Not quite, I had assumed the same, but when I drew out my representative chassis, what becomes clear is that the caster angle is originated at the front tyre to ground contact point, whereas the frame angle is originated at the front wheel hub, so the two angles are not directly linked as they have different origin points. In a way it’s similar to unequal length wishbone suspension, where the wheel moves the ball joints the same distance but because the arm to pivot lengths are different they end up at different angles.

 

[Righty]

I reckon I might have to get the Lego Technic out and build a model

 

[Woody500]

Yes I agree, dividing the 1.7 by 108 does give that figure, which as you say equates to a tiny angle change. Thats the reason i ended up plotting it out in my CAD programme (I design exhibition stands), because when I put the ratio of V 1.7 (43mm) to H 108 (2743mm) into an online calculator it returned the degree value of 0.898.

View attachment 1716436429


So I drew up a very basic representation of the frame rails at horizontal and sitting at the current rear height, with a representative upper BJ set horizontally back from the contact point of the tyre to give a nominal caster of 3 degrees. Then using the centre of the front wheel as the pivot point I rotated the frame to reach the new height at the back +1.7" so the dashed horizontal line on the second drawing is the original line of the frame rail and the angled dashed line represents the frame angle with the rear lifted +1.7 the dimensioning tools in the CAD programme are generated by the programme not my maths so it automatically finds the angle between those 2 lines.

Also another thing I wasnt expecting, I dont know if you saw on the drawing, but the change in frame angle does not directly corelate to the reduction in caster, so the frame angle is +0.88deg but this only reduces caster by -0.29deg. I had assumed that it would be like tilting a table and the top and the legs would rotate by the same amount. In this case I believe that the reason it doesnt is that the caster angle is from the contact patch to the upper BJ whereas the frame angle is from the rear of the wheelbase to the centre of the front wheel so the origin points of those 2 angles are in different places creating 2 differntly proportioned triangles.

It was interesting Blu's comment above seems to tally in roughly with what I found from the drawing.


Anyway for all the drawings and calculators, the reality is I need to drop the rear height using the adjustable spring hanger because it looks silly and isnt as nice to drive as before. I doubt ESPO sent the incorrect springs their reputation tends to make me think not, but the springs that were on there must have been really worn out as the difference in height now is pretty big. To be honest going out and re checking from axle centre to arch with the wheel off its probably more like 2.5". Lucky I fitted adjustable hangers!

EDIT: SO I THINK I HAVE THE ANSWER!

SOH CAH TOA. sine would be for if i knew the opposite and hypoteneuse sides, I dont, I know the opposite and adjacent so I need tan, actually I need inverse tan. School was worth something.

View attachment 1716436430

Espo shouldn't charge extra for lowered springs. And I'd bet they sell plenty of them, & it costs NO extra to do them that way.
Unless , of course, they get large lots of pre arched stock, from 'Somewhere' , and then assemble & re-arch for a mark up.
So no sale here.
Anyone ever checked what alloys they are with a metal scanner?

 

[Killer6]

So I just checked the suspension height at the front, from the bottom of the adjustment blade to ground is 8.7" and from the bottom of the LCA knuckle is also 8.7". According to the FSM the suspension height for the Barracuda should be 1 3/8" +- 1/8". So unless i am taking my measurements in completely the wrong place that would be 1 3/8 lower than factory spec interesting.

When performing alignments on a lot of Chebby trucks, you have to go thru' a "rake calibration" by direct measurement & clicking 'up/down' on the scale image to get the figure to match Your actual readings as instructed, this is specifically to automatically adjust what the correct caster should be. YES, raising the rear reduces caster, dropping the rear the inverse.