# Axle measurement help...Moser screwed up

#### xoxbxfx

Founding Member
Ok guys I need help...here is the drawing of the axles I have now now stock 99+ axles have the following measure:
diag: 30 11/16"
axle shaft length: ???

I needed axles with 2" shorter axle shafts, not diagonally like Moser likes to measure. I know a^2 + b^2 = c^2 but Im getting the hub RADIUS of 8.13". Its a right triangle with the longside being 28 11/16", the bottom as 27.5" and that makes the radius 8.13" unless Im doing my math incorrect.

I told the guy on the phone when ordering axles I narrowed the rear 4", 2" on each side. I needed axles with 2" shorter axle length, not diagonal length....apparently he did not understand. Apparently NO ONE understands up at moser when you modify a rear end and cut out the length from the housing you are altering the axle shaft length thats its not the same as taking out 2" of the diagonal length.

I need the diagonal length of a stock 99" mustang 8.8 axle that has been shortened 2" in the axle length, not diagonal length. They stand by their product saying that they are correct...I dont understand how they come up with their math.

in a 3,4,5 triangle the math works like this... 3^2 + 4^2 = 5^2 which equals 9+16=25.... now with their math they say 3^2 + 3.5 ^2 = 4.5^2... 9 + 12.25 does not equal 20.25... that 2nd equation would be me saying I took out .5" of the bottom side and their math says it equals .5" out of the long side. They are 1 full inch off on their measurement....

Can anyone help?

Your pathagorean (SP?) theorum formula is right..However, in your diagram, the height (wheel hub diameter as labled) might have to be 1/2 the wheel hub diameter? It looks like the lenght horizontally is the axle itself, with the diagonal being from the center of the axle at its tip/the end of the axle. This would mean that to find the diagonal, you would use 1/2 the hub radius. (you labled it "????"). Just an idea.

Your pathagorean (SP?) theorum formula is right..However, in your diagram, the height (wheel hub diameter as labled) might have to be 1/2 the wheel hub diameter? It looks like the lenght horizontally is the axle itself, with the diagonal being from the center of the axle at its tip/the end of the axle. This would mean that to find the diagonal, you would use 1/2 the hub radius. (you labled it "????"). Just an idea.

thats the problem... pythagorean theorum only applies to right triangles... the right triangle in this case would be 1/2 (length wise) through the center of the axle to the top of the hub. If the 8" were the whole hub, the triangle length would shift to the bottom of the hub and all the angles would change

Well, I think I figured it out after I measured the stock axles...

Axle length = 30.54051
Diag length = 30.6875
Hub diamter = 6"

What I have according to the same formulas with a 2" narrower diagonal measurement:
Axle length = 28.53021
Diag length = 28.6875 (2" narrower")
Hub diameter = 6"

What I need by subtracting 2" from the axle length of the stock axle:
Axle length = 28.54041
Diag length = 28.67775
Hub diameter = 6"

so all in all, its off by about 1/100th of an inch... wtf, guess the diag length is close enough to the axle length that it doesnt matter