Hey there. Ive been doing some checking around and have found it there way only two rear end ratios availabe in 66. A 3.50 and a 4.00. I decoded my VIN to try to figure out what I have, but according to the vin I have a 68 3.50 rear axle. The numbers for 66 simply arent there on the VIN tag, just the 68 numbers. I am confused and cannot figure out which rear end I have. Is there another way to figure out?


Thanks


Johnathon

 

Don't know where you looked but there are 29 different axle codes for a 1966 and about 11 different ratios... http://www.home.earthlink.net/~jcpurut/vin/vin.html

You should/will also be able to find a tag on your differential. My guess is you have a 9" Ford and the metal tag would hang from the front where the chunk goes in. What code do you have on your VIN Tag?

--Mike

 

I Gotta agree with Mike.

There were far more than 2 ratios available in 66 for 2 whel Drive, F Series with 9" FoMoCO Rear axle assy. Perhaps you mean there were only 2 types of rear ends available?

The 9" FoMoCo hotchkiss type with front loading third member was the more common & a Dana 40, integral housing, rear loading Salisbury type was the less common, and usually only came with the 240 CID I-6.

Dana only had 2 ratios, but I think they were 4.10:1 and 3.73:1 respectively.

I see 9 ratios for the 9" and the 2 ratios I already mentioned for the Dana. And that's what Mike said. . . . 11 options in 66.

You can jack up one rear wheel and support the vehicle with a jackstand, block the front wheels in 2 directions, put the trans in neutral mark your driveshaft with a piece of masking tape, mark a tire with another piece. Make some kind of index for accuracy so you can see how many actual revolutions or parts of a rev' the driveshaft makes.

Turn tire 1 revolution and count how many revolutions your driveshaft makes.

This is not as accurate as actually counting teeth & doing the math, but it will get you close enough so you have some idea of what your gear ratio is.
3 turns = 3.00, 3 & 1/4 = 3.25, 3-1/2 = 3.50, 3-3/4 (mol)= 3.73, so forth & so on.

Also I never heard of nor seen a 4.0:1 rear end in a Ford. 4.10, 4.27, 4.56 but no 4.0
You might want to recheck the sources that gave you info you have. It's not correct.

Good Luck

FBp

 

Heaven forbid me to correct the venerable FordBoyPete -- but he himself will note his error surely. In an open rear, with one tire down on the ground as he suggests, the axel will turn only 1/2 the number of revolutions as the ratio, per revolution of the tire.

That's becuase of the differential turning the free tire twice as fast. So:

1.5 axel rev = 3.00 rear, 1.75 axel rev = 3.50 rear etc.

Best thing to do is rotate the tire (marked as Pete suggests) 2 revolutions -- then the number of rotations of the axel will be the correct ratio of the rear.

Or, if you desire greater accuracy, such as finding out if its a 3.5 versus a 3.25 (which is tricky), then tape a string or thread to the drive shaft and rotate the tire 10 times. Count the number of times the thread is wound up on the drive shaft, then divide by 5.

 

Send me the info. You probably have a 3 to 90 ratio.

 

OOOooops! I stand corrected. Thanx CD that's what friends are for, and I was way off there,what was I thinking. Completely left spider gears out of equation on that one.

Now if I can just get me foot out o' me mouth here. . .to quote Homer Simpson DOoH..

BTW Scottish Mannish ? ? ? ?
What do you mean by a 3 to 90 ratio?
That's a new one to me. 3 : 90 = 1 : 30 in corrected form, or lowest denominator value. Do you mean a 3.90:1 ratio? Which I believe will actually be a 3.89:1 if it's a Ford gear.

Thanx again for catching that blunder, "CD", I preciate it.

FBp