Today rather than reinvent the wheel, my question is more along the lines of working around with the wheel... Lets see if I can formulate my question properly.

My '56 has a 1978 351W, C6 Tranny (off of the same donor - a E150 Van) and the "original" '56 rear end --which according to the V.I.N. plate it is a 3.92. I like the way this is going so far.

So armed with this information I head on down to my local truck shop which specialized in differentials, gear swapping and all kinds of goods stuff to find out if this combination is good for attaining "cruising" speeds (anywhere from 55 to 70 MPH) and they stomp me with this question - but keep in mind, I'm easily stompable!

What size tires are on the truck? 15's I proudly reply. He smiles and says, "No, the height of the tire" Huh? Double huh, huh? F^(K!! He goes on to educate me by stating that in order to attain what I desire, one must consider this very small - yet big - detail. Evidently, it matters a lot.

So rather than trying to make all kinds of changes in rear end gearing - for now - my question is this one:

Since my configuration is as mentioned above, what would be the ideal (as close to as possible) tire size so that I can "cruise" - even if not at 70 MPH but at least 55 MPH? The goal is to keep the RPM around 1800 to 2200 while at the speeds mentioned.

I have looked at the gear ratio calculator website - Gear Ratio Calculator - but quite honestly, I can not seem to make it work for me, most likely because I do not know some of the info it requires.

I'll check back a little later - gone for now... Thanks in advance!

Let's start this discussion literally where the tire meets the road. We need to know how far the tire rolls per revolution; so, let's assume that you have a 30 inch tall tire. The circumference of this tire is 3.14 X 30 or about 94 inches. So now we know that in 1 rotation of the axle the truck will roll 94 inches down the road.

That's nice to know but how does that relate to engine RPM you ask. OK, there are 5,280 ft in a mile which makes it 63,360 inches/mile. At a speed of 70 MPH you will be traveling 1.17 miles in a minute or 73,920 inches/minute. What we really want to know is how many revolutions of the tire will occur at this speed so divide the 73,920 by 94 and we find that the tire must rotate at 786 RPM.

Moving along, the tire rotates at the same rate as the axle but the axle rotates at a different rate than the driveshaft. With a 3.92 gear ratio in the differential we can calculate the driveshaft rotation needed to give us the 786 RPM at the axle by multiplying 786 by 3.92 which equals a driveshaft RPM of 3,081.

OK so now we know that a truck with a 3.92 rearend and 30 inch tall tires traveling at 70 MPH will have a driveshaft rotating at 3,081 RPM. How does this relate to engine RPM? Well that is where the transmission data comes into play. Most automatic transmissions have around a 1:1 drive ratio...that makes our engine calculation very easy. However, if you are running an overdrive then the calculation gets a little trickier. I have an AOD with a 0.62 overdrive ratio so with a driveshaft RPM of 3,081, my engine will need to turn at 3081 X 0.62; or at 1,910 RPM.

Let's start this discussion literally where the tire meets the road. We need to know how far the tire rolls per revolution; so, let's assume that you have a 30 inch tall tire. The circumference of this tire is 3.14 X 30 or about 94 inches. So now we know that in 1 rotation of the axle the truck will roll 94 inches down the road.

That's nice to know but how does that relate to engine RPM you ask. OK, there are 5,280 ft in a mile which makes it 63,360 inches/mile. At a speed of 70 MPH you will be traveling 1.17 miles in a minute or 73,920 inches/minute. What we really want to know is how many revolutions of the tire will occur at this speed so divide the 73,920 by 94 and we find that the tire must rotate at 786 RPM.

Moving along, the tire rotates at the same rate as the axle but the axle rotates at a different rate than the driveshaft. With a 3.92 gear ratio in the differential we can calculate the driveshaft rotation needed to give us the 786 RPM at the axle by multiplying 786 by 3.92 which equals a driveshaft RPM of 3,081.

OK so now we now that a truck with a 3.92 rearend and 30 inch tall tires traveling at 70 MPH will have a driveshaft rotating at 3,081 RPM. How does this relate to engine RPM? Well that is where the transmission data comes into play. Most automatic transmissions have around a 1:1 drive ratio...that makes our engine calculation very easy. However, if you are running an overdrive then the calculation gets a little trickier. I have an AOD with a 0.62 overdrive ratio so with a driveshaft RPM of 3,081, my engine will need to turn at 3081 X 0.62; or at 1,910 RPM.

I hope that this helps more than confuses....

Well said. So easy even I can understand it. You should be a teacher.

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So many variables! I don't know how many gears a C6 has, but I can tell you with my 3 speed tranny (and I don't even know what the ring and pinion are in my truck), that it will cruise at 55 mph, but not go much over 65mph. The speed limit on the highways in Colorado is 75mph, but obviously most people are going 85. I can't even venture out there with my truck. I think the best approach is to get a 4 spd trans, that makes it much more easy to obtain lower RPM highway speeds without losing acceleration. Then IF you need to, replace the rear-end gears. I know when this tranny goes I'm getting a 4 or 5 spd trans to replace it.

So many variables! I don't know how many gears a C6 has, but I can tell you with my 3 speed tranny (and I don't even know what the ring and pinion are in my truck), that it will cruise at 55 mph, but not go much over 65mph. The speed limit on the highways in Colorado is 75mph, but obviously most people are going 85. I can't even venture out there with my truck. I think the best approach is to get a 4 spd trans, that makes it much more easy to obtain lower RPM highway speeds without losing acceleration. Then IF you need to, replace the rear-end gears. I know when this tranny goes I'm getting a 4 or 5 spd trans to replace it.

You are really confused...no matter how many gears you have or whether it is an automatic or a manual shift, the final gear ratio is what matters. Virtually all non-OD transmissions are 1:1 in high gear. Most of these old trucks have low gearing because they were built to haul loads on secondary roads, not modern day highways. If you want to make your truck perform better on the highway take a look at your differential, not your transmission...that is unless you want to add an overdrive.

Let's start this discussion literally where the tire meets the road. We need to know how far the tire rolls per revolution; so, let's assume that you have a 30 inch tall tire. The circumference of this tire is 3.14 X 30 or about 94 inches. So now we know that in 1 rotation of the axle the truck will roll 94 inches down the road.

That's nice to know but how does that relate to engine RPM you ask. OK, there are 5,280 ft in a mile which makes it 63,360 inches/mile. At a speed of 70 MPH you will be traveling 1.17 miles in a minute or 73,920 inches/minute. What we really want to know is how many revolutions of the tire will occur at this speed so divide the 73,920 by 94 and we find that the tire must rotate at 786 RPM.

Moving along, the tire rotates at the same rate as the axle but the axle rotates at a different rate than the driveshaft. With a 3.92 gear ratio in the differential we can calculate the driveshaft rotation needed to give us the 786 RPM at the axle by multiplying 786 by 3.92 which equals a driveshaft RPM of 3,081.

OK so now we know that a truck with a 3.92 rearend and 30 inch tall tires traveling at 70 MPH will have a driveshaft rotating at 3,081 RPM. How does this relate to engine RPM? Well that is where the transmission data comes into play. Most automatic transmissions have around a 1:1 drive ratio...that makes our engine calculation very easy. However, if you are running an overdrive then the calculation gets a little trickier. I have an AOD with a 0.62 overdrive ratio so with a driveshaft RPM of 3,081, my engine will need to turn at 3081 X 0.62; or at 1,910 RPM.

I hope that this helps more than confuses....

To answer your original question more directly...assuming that you are in fact running a 30 inch tall tire then to get an engine RPM of 2000 you would use the formula 786 x rearend ratio = 2000. This calculation returns a ratio of 2.55. It is noteworthy to add that the taller the tire the lower the axle RPM. For example, the 786 RPM I calculated for a 30 inch tire would become (73920/(3.14x32)) = 736 RPM for a 32 inch tire. Soooo, with a 32 inch tire at 70 MPH and an engine RPM of 2000 you would need a gear ratio of 2000/736 = 2.72

Hopefully this is enough info so that you can do your own calculations once you have measured your tire height.

Thank you so much!!! I will need to re-read the info you so eloquently provide. Then, I will arm myself and visit my local truck center again and at least sound intelligent. Honestly, it DOES make sense!

CharlieLed thank you so much and to the others who chimed in thank you as well. Thank you all.

This weekend I will start re-assembling the engine and tranny. Will post pictures once progress is made. Stay tuned.

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