Dylanamus

1990 CR-X SiR - EF8 with factory b16a and a whole heap of annoyingly rare and expensive parts.




The car is lowered on fully adjustable coilovers. Obviously one side is not lowered equally with the other, as demonstrated by the camber specs.

The problem here is the rear toe. I can't figure it out! If the compensator arm bolts were position in the centre of the subrame mount where the factory alignment should be at zero, the toe would be at like -15mm per side. Ridiculous!

I couldn't see any signs of bent components, so I'm guessing at some stage the wrong sized trailing or compensator arms were installed?

Thoughts?


I am thinking adjustable compensator arms is the solution. Is +10mm/-10mm adjustability available in the aftermarket? Answer is yes, according to the below product.


Most toe kits show the adjustment range in degrees, so I used this formula:

1. Multiply rim diameter in inches by 25.4 to get rim diameter in mm.
2. Multiply the result by PI to get the circumference.
3. Divide the result by 360 to get mm per degree.
4. Divide the toe mm by the mm/degree result to get toe in decimal degrees


15" x 25.4 = 381mm
381 x pi 3.14159265 = 1196.95
1196.95 / 360 = 3.325

I'll use an ebay kit as an example (as below)

Adjustment range: -2.00 degrees to +5.00 degrees
Adjustment range: -6.65 mm to +16.6 mm


[url=http://cgi.ebay.com/ebaymotors/WICKED-REAR-TOE-CAMBER-KIT-HONDA-CIVIC-CRX-EF-88-91-S_W0QQcmdZViewItemQQ_trksidZp1713Q2em153Q2el1262QQ categoryZ33581QQihZ007QQitemZ170207227594QQtcZphot o[/url]


Modified by Dylanamus at 1:42 AM 4/2/2008


Modified by Dylanamus at 5:21 AM 4/2/2008


Modified by Dylanamus at 7:07 AM 4/2/2008

IT GUY

those mfactory arms you have the link to are nice. don't buy them off ebay, buy from mfactory on honda-tech (check marketplace)

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by dylanamus &raquo;</TD></TR><TR><TD CLASS="quote">
Most toe kits show the adjustment range in degrees, so I used this formula:

1. Multiply rim diameter in inches by 25.4 to get rim diameter in mm.
2. Multiply the result by PI to get the circumference.
3. Divide the result by 360 to get mm per degree.
4. Divide the toe mm by the mm/degree result to get toe in decimal degrees


15" x 25.4 = 381mm
381 x pi 3.14159265 = 1196.95
1196.95 / 360 = 3.325 </TD></TR></TABLE>

I'm not following this formula entirely and I can't find in the ebay ad the adjustment range you used. Leaving out the tires?

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by MKIV.com &raquo;</TD></TR><TR><TD CLASS="quote"> To Calculate Degrees of Toe When Given mm of Toe:
*Note that this is dependent upon the wheel diameter and tire diameter

Formula: Calculate the Wheel + Tire Diameter (in mm) for the stock rear wheel + Tire
Example: (2 * tire width * aspect ratio / 100) + (wheel diameter) = wheel + tire diameter
(2 * 255mm * .4) + (17 * 25.4) = 635.8 mm

Here the goal is 1 mm of total toe in. (This is 0.5 mm on the driver rear tire and 0.5 mm on the passenger rear tire.)

We want to calculate the angle each side wheel + tire must deviate from a line parallel to the centerline of the car.
arcsin(0.5 mm / 635.8 mm) = .045 deg
.045 deg per side * 2 sides = .09 deg total toe in </TD></TR></TABLE>

Dylanamus

Tyres! Completely forgot.. ummmmm

I can't do arcsin! I was never good at math, but now I don't even have a calculator with arcsin on it. So how do I reverse the calculation to use SIN only?

Tyres are 195/50R, so (2 * 195 * .5) + (15 * 25.4) =

390 + 381 = 771

arcsin(0.5 mm / 771 mm) = 2.295 deg ??

arcsin(1 mm / 771 mm) = 3.5388 deg ???

PS: I redirected the above URL to the correct link. It wasn't supposed to be mfactory! It would seem that degrees is a useless way to market the adjustability. What variables are they basing this data on? Who knows.... I guess to even begin making it relevant, I really need to calculate the degrees of my current rear toe, to then determine how far from zero degrees it is and then figure it out that way. Although this still asks the question of the manufacturer, degrees? Based on what?


Modified by Dylanamus at 7:10 AM 4/2/2008

harlen

1/sin(x) = arcsin(x)

Dylanamus

Going back to roots...


Step 1. Calculate the Radius (in mm) of the wheel and tyre together.

<FONT COLOR=""Red"">Formula</FONT> (2 * tire width * aspect ratio / 100) + (wheel diameter) = wheel + tyre diameter

<FONT COLOR=""Blue"">Executing first equation:</FONT>

Wheels are 15" and tyres are 195/50R, so (2 * 195 * .5) + (15 * 25.4) =
195 + 381 = 576

576/2 = RADIUS = <FONT COLOR=""#ff0000"">288</FONT>

Step 2. Calculate the angle each side wheel + tire: multiply the sin of 1 degree by the radius of the rim + tyre, to give you the mm deviation from parallel with the centreline (CL)

<FONT COLOR=""Red"">Formula</FONT> arcsin(radius) = radians per mm

<FONT COLOR=""#0000ff"">Executing second equation:</FONT>
sin(1) (degree) or sin(pi/180) (radian) = 0.017452

0.017452 X 288 = 5.026293mm

<FONT COLOR=""Purple"">Conclusion</FONT> For my wheel/tyre combo every 5.026293 equals one degree.





I'll use an ebay kit as an example (as below)

Adjustment range: <FONT COLOR=""Blue"">-2.00 degrees</FONT> to <FONT COLOR=""#0000ff"">+5.00 degrees</FONT>
Adjustment range:<FONT COLOR=""Red"">-10.05 mm</FONT> to <FONT COLOR=""#ff0000"">+25.13 mm</FONT>


Modified by Dylanamus at 5:03 AM 4/3/2008

Dylanamus

Question for the math or mech engineering nerds.. the point on which the compensator arm connects with the trailing arm is a fair distance beyond the wheel; does this affect the accuracy of my calculation much?