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Spring Plate Angle Calculator - Beta

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Old Jan 3, 2002 | 10:21 PM
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Post Spring Plate Angle Calculator - Beta

For your review - thanks to Will Ferch for the actual math.

Spring Plate Angle Calculator
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Old Jan 3, 2002 | 11:28 PM
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Did you use it when you put the new t-bars in? How'd that all go?
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Old Jan 4, 2002 | 12:50 AM
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I didn't use it to calculate my own spring plate angles, but it's eerily dead-on to the angle I came up with on my own. I'd been having a discussion with Will Ferch about his mathematical means of arriving at spring plate angles, and the programmer in me sprang out... FWIW, it's right-on for my spring plate angles before and after the swap.

As far as the actual process, it went OK 'cept for a wee bit of a broken sway bar mount. I spent 16 hours last Thursday fixing the mount and getting the bars installed.

I wrote up the procedure at: http://www.vintagebus.com/howto/torsion

Originally posted by Bill Verburg:
<STRONG>Did you use it when you put the new t-bars in? How'd that all go?</STRONG>
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Old Jan 4, 2002 | 01:57 PM
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I suck at remembering things so I often have to derive the formula every time. If you can remember that torsion bar stiffness increases at the 4th power of its diameter (twice as thick = 16 times as stiff) and that a spring compresses a distance proportional to its stiffness (twice as stiff compresses half as much) you can derive the formula any time you need it.
Example: If Sport spring "A" is twice as stiff as Stock spring "B" it will only compress half as much. In our case we're dealing with torsion bars so substitute angle for distance.

new angle = old angle / stiffness change

angle: the angular difference between the spring plate angle in compressed (car on the ground) and noncompressed states (car in the air)

stiffness change: torsion bars increase in stiffness at the 4th power of their diameter so the stiffness change would be:

stiffness change = (new dia / old dia) ^ 4

For the following example I'll use:
old bar 25mm thick
new bar 29mm thick
The new bar will be 1.16 (29/25) times as big as the old one.

stiffness change = 1.81 times as stiff(1.81 = 1.16 x 1.16 x 1.16 x 1.16)

(Come back! come back! I'm almost done!)

Soooo, if the old angular difference between the compressed and uncompressed spring plate was 10 degrees we'd want to put the spring plate on at a new (smaller) angle

old angle = 10 degrees
new angle = old angle / stiffness change
new angle = 10 deg / 1.81
new angle = 5.2 deg

I use some wide masking tape on the inner fender and scribe a line on it along the top of the spring plate while the car is sitting on the ground. I'll call this the "reference line"

Jack up the car and measure the angle between the now drooping spring plate and the reference line. This is the old angle.

Using the formulas we derived make a template out of cardboard with the new angle and use it for both sides when you reinstall the spring plates with your new bars.

Of course if you take this opportunity to replace your hydraulic shocks with new gas shocks all your measurements will be for naught because the gas shocks act like springs.

-Chris
"Lot of time on your hands Chris?"
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Old Jan 4, 2002 | 03:12 PM
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Chris:
If only it were so easy to do...then this wouldn't have been the problem it was for 30 years to establish proper torsion bar resetting for the 911 pattern car. For example, a 20 degree twist of a torsion bar results in different ride heights of you go from 4:00 o'clock to 2:00 O'clock position...compared to going from 7:00 o'clock to 5:00 O'clock , where there wouldn't be a height change at all ( visualize this). This is all described in the tech section where I posted on this topic a long time ago. Thom was good enough to work with me ( and his own situation)...to work up a quick software data input version that collapses all the hard math into a few easy entries.
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Old Jan 4, 2002 | 09:40 PM
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Originally posted by Wil Ferch:
<STRONG>Chris:
If only it were so easy to do...then this wouldn't have been the problem it was for 30 years to establish proper torsion bar resetting for the 911 pattern car. For example, a 20 degree twist of a torsion bar results in different ride heights of you go from 4:00 o'clock to 2:00 O'clock position...compared to going from 7:00 o'clock to 5:00 O'clock , where there wouldn't be a height change at all ( visualize this). This is all described in the tech section where I posted on this topic a long time ago. Thom was good enough to work with me ( and his own situation)...to work up a quick software data input version that collapses all the hard math into a few easy entries.</STRONG>
Will,
I looked into this a few years back and started off by considering the changing rate depending on the angle i.e. the effective moment arm is a different depending on the angle of spring plate. Then I had an epiphany The angle of the spring plate is irrelevant/drops out because the spring plate is returned to the original angle (ride height.) We don't need to figure in how much the ride height changes based on spring plate rotation because we aren't changing the ride height.
In other words, my premise is that all we care about is the forces applied when the spring plate is at the compressed angle. Since the angle before and after (and thus the moment arms) are the same the angle itself doesn't matter.
Chris
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Old Jan 8, 2002 | 12:50 PM
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Chris:
I'm not sure exactly what you're saying. Let me be a bit more clear. First, you need to read my article in the Tech section that was the jumping-off point for the work Thom Fitzpatrick did. In it, you will note the inter-relationship of torsion bar stiffness (new big vs old small)..the free hanging angle, and the spring plate angle you end-up with at static ride height. The spring "rate" will be known in any given circumstance. I'm simply saying that recognizing the "fourth power" rule ( only) won't get you there.
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