Formula to calculate engine RPM from speed, tire size, gear ratio, final drive, and estimate fuel economy

Power/Fuel
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No need to go no the Internet and have a website with advertisements and Javascript for a simple formula that you can write down. Any high school scientific calculator can do this without the need to do the order of operations by hand.

The first formula for wheel circumference:
Where W D and R correspond to tires size like [width in mm] [depth %] [rim in inches]

Wheel circumference (in) = (W x D / 1270 + R) x 3.14

Example: 195/70 r14 tire is (195 x 70 / 1270 + 14) x 3.14 = 77.7 inch circumference.

Multiply this by .97 to get the squish factor. 77.7" is now 75.37.

Speed formula:
E = engine RPM
R = transmission ratio
F = final drive ratio
C = wheel circumference inches from previous formula
E / (R x F) x C / 1056 = speed in MPH

Example: 2200 RPM, 0.8 transmission ratio, 3.55 final drive, 235/75r15 tires (that’s 87.5" circumference):
2200 / (0.8 x 3.55) x 87.5 / 1056 = 64.2 MPH

To find the engine RPM from speed, reorder the formula like this:
S = Speed in MPH.
S x R x F x 1056 / C = E

I would like to tweak the wheel circumference formula. I would like to have a more advanced version that takes in to account tread wear, weight, air pressure, or at least general real world adjustment coefficient. Any suggestions are welcome. There doesn’t seem to already be a discussion on this.

I got started with this because the 2767lb 2020 Kia Rio 1.6L CVT gets EPA 41 MPG highway, but 2714lb 2019 Kia Rio 1.6L 6 speed manual gets 37 MPG highway. The 6 speed’s lowest ratio is 0.703 with 4.267 final drive, which is 3.00 total. The CVT ranges between 2.68 to 0.39, so 0.39 with its 5.45 final drive is 2.13 total. That’s a big difference.

If the 2019 Kia Rio’s engine is at 2000 RPM in 6th gear with its 185/65r15 tires (76.8" circumference), it would be going only 48.5 MPH.

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That should be 77.7 inches. A fat-fingered mistake I’d guess.

Looking up revs/mile for that size tire on Tire Rack Specs, you get 839 revs/mile for a 24.7 tall tire in the size you chose as an example. That would seem to be a 77.6 inch circumference - brand build variations so pretty close.

Rolling diameter = 63,360 / 839 = 75.5 inches rolling circumference, or 97% of your calculated value. Static loaded radius will always be smaller than any calculated size.

You need a 3% adjustment since the tire sags a bit even if properly inflated. That 3% will likely vary between aspect ratios… but I have not checked…

Edit;: Just checked a 275/30/19 tire… the 3% squish factor is still valid!

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And payload, driver input as far as acceleration and braking, tire rolling resistance, pavement surface, rain or shine, parasitic engine loss from electrical loads like defroster and A/C, etc…

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I have never felt the need to calculate engine RPM, but then I do not do cryptograms, wordle, or math exercises. I think about other things, like why hasn’t Wile E Coyote ever caught the Roadrunner or why isn’t it obvious to Lois that Clark Kent is Superman?

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… or why did Clark never get into trouble with the cops when he used outdoor telephone booths to change into his Superman costume?

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Fuel mileage is not just a factor of engine rpm. Rolling friction matters a bit, aerodynamic effects dominate above about 40 mph.

Aero is a factor of frontal area, the coefficient of drag and speed.

Drag = 1/2* air density * frontal area * speed^2 * drag coefficient

Twice the speed creates 4 times the drag.

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It’s 8 times the drag, but 4 times the energy used, since you get there in half the time, half as much of the 8 times as much energy is needed.

I think when I’m going 52 MPH, at least half of my engine’s fuel is used to keep the engine 2000 RPM, and the rest is actually used to produce usable energy so I get say 30 MPG. At 1500 RPM, fuel use would be reduced down to 3/4 in an extreme theoretical example, so it would be 40 MPG. In this case only the wasted fuel would be reduced to 3/4, and if the wasted fuel is half, then that’s really down to 7/8 consumption, so it would be more like an improvement up to 34 MPG.

Why is it necessary to talk about hating math and Road Runner and then go on talking about that? I’d like to stay on topic.

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Do you see a cubic function there? No. Drag is a square function.

Horsepower to reach that speed or maximum speed is a cubic function.

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I live in Albuquerque, where roadrunners are common. I’ve nearly run them over on my bicycle: they aren’t that fast. You can find real encounters between the 2 on YouTube: guess who wins.

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So drag must mean the force that must be overcome to move through the air. 8 times the energy is needed to go twice as fast, the force pushing against is 4 times.

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Drag is a force. Energy is the fuel in the tank waiting to be used. Power is the rate of use of that energy. That is a cubic function with respect to aero drag.

Power to push the car down the road is rolling friction plus aero plus any gradient load times drivetrain losses. Rolling friction is roughly linear as is drivetrain efficiency.

Fuel used to make that power depends on many more things that can be distilled down to the efficiency of the engine and driveline to apply that power to the rear wheels.

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We had a few roadrunners in the Mojave, though many more coyotes than road runners.