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For you mathematicians out there, I have a Ruckstell of unknown ratio. How do I determine that?

Also, given that ratio and standard size wheels & tires, how do I determine engine RPM in high gear at 35MPH?

Thanks! Bill

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jack the car up pull the plugs turn the crank until the rear wheel goes i rev.

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Diameter of tire (30 inches) times pi = 94.25 inches per revolution, or 7.85 feet. There are 5,280 feet in a mile so the wheel revolves 672 times in a mile. So if you have a 3.63 ratio rear end the engine will turn 3.63 X 672 = 2,440 revolutions in a mile. If you are driving at 35 mph that would be 35/60 X 2,440 or 1,424 rpm. At 14,240 rpm you would be going 350 mph and your engine would be in several different counties.

If you jack up one wheel the wheel will turn twice as much as if both wheels are jacked up. It is easier and more accurate to count the engine revolutions while watching the one jacked up tire turn 10 revolutions. You then divide the number of revolutions by 5 to get the gear ratio. So if the engine turns 18 revolutions you have an 18/5 = 3.63 to 1 gear ratio. Also easier if you take the plugs out while you crank the engine.

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Or leave on the ground take out plugs with the car in high/direct push the car forward one rev of the wheel and count how many times the engine turns over. Standard you should get about 3 and 1/2+.

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The Fast Ford handbook lists 400 rpm per 10 mph with stock gears, 333 rpm per 10 mph with 3:1 gears.

Andy

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I would not go by the tire diameter since the bottom of the tire under load is compressed. The “effective” diameter is determined by measuring the height from the ground to the center point of your hubcap (rolling radius) and multiplying that by 2. If a 30” tire is compressed ¼” under load, the effective diameter is 29 ½”. To simplify Neil’s calculation you can use this:

RPM = MPH x Axle Ratio x 168.07 / Rolling Radius

Example:

35MPH x 3.63 x 168.07 / 14 ¾” = 1,447 RPM

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A 1922 chart from Ford for the standard ratio,

Ken in Texas

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Mark,

Although the bottom of the tire is compressed, it doesn't change the circumference. There is likely some deformation, though. Probably best to mark the tire and floor/driveway/street, roll exactly one revolution, and measure distance traveled.

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If you have the differential ratio, or any three of four, rpm, diff ratio, speed and tire size this is a good calculator. I've found our "squish" of the tires lowers the tire diameter about a 1/2 inch under actual tire size.

http://www.advanced-ev.com/Calculators/TireSize/

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OK, I stand corrected. I really wouldn't have thought it could be that much. That equates to better than 1-1/2" on the circumference.

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The squish does not effect the circumference. An example would be a drive belt. A 30 inch belt is 30 inches no matter how many pulleys you wrap it around. The radius of the tire at the ground will effect the leverage, or effective gear ratio. It is easier to calculate the circumference with a constant radius, but the 30 inch tire will always have a circumference of just over 94 inches

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And here I thought Neil's explanation (at 12:10 PM) had a very complete, thorough explanation!

"Happy T-ing!"

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I got a real kick out of Neil's "Bonneville" calculation! Thanks to all. I'm researching crankshaft seals and an important spec is how many RPMs the shaft is spinning. The packing that looks good is rated at 2,500 RPMs which should be plenty. I drive low and slow. I will start figuring. Many thanks to all! Bill

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I remember several years ago the late RDR posted a a chart with rpms = to MPH. Using stock ratio and 3/1 , Used to have it on my old outer. Wish I had it now so I won't over rev.