I’ve been thinking about the effect of larger tires (weight and diameter) on a vehicle’s torque and HP to the wheels. I wanted to calculate some “real world” numbers that would identify the actual effect that increasing tire size would have on my vehicle. I came across the following formula on a physics forum that calculates the energy required to rotate one wheel, based solely on weight, to a speed of an arbitrary 60 MPH.
"Suppose the moment of inertia of each wheel (rim plus tire) is I = (3/4) m R2
where the 3/4 is a coefficient related to how the mass is distributed radially, and R is the rolling radius.
The energy of one wheel rotating at a velocity of w (in radians per sec) is
E = (1/2) I w2 = (3/8) m (Rw)^2
But Rw is just the velocity of the rolling wheel with radius R, so Rw= v (meters per sec). So for one tire, the energy is E = (3/8) m v2
For a wheel (rim plus tire) mass of X pounds (X Kg) and 60 mph (26.8 meters per sec)
E = (3/8) (X) (26.82^2) = Y Joules
1 joule = 1 watt
1 watt = 0.00134102209 horse power
If we had to accelerate four wheels from 0 to 60 mph in 4 seconds, the power would be P = 4 x Y Joules/4 sec = Z watts = HP, INDEPENDENT of rolling radius!"
Reference https://www.physicsforums.com/threads/calculating-torque-hp-losses-from-wheel-choices.334984/
I used the specs for the BFG AT KO2 in several sizes and plugged them into the formula above for comparison sake. I came up with the following:
LT265/70R17/C 31.6 “ 45.7 lbs (20.73 kg)
E = (3/8) (20.73) (26.82^2) = 5596 Joules
5596 / 0.00134102209 = 4.17 HP
LT265/70R17/E 31.6” 53.4 lbs (24.22 kg)
E = (3/8) (24.22) (26.82^2) = 6533 Joules
6533 / 0.00134102209 = 4.87 HP
LT275/70R17/E 32.2” 55.4 lbs (25.13 kg)
E = (3/8) (25.13) (26.82^2) = 6779 Joules
6779 / 0.00134102209 = 5.08 HP
LT285/70R17/E 32.7” 58.1 lbs (26.35 kg)
E = (3/8) (26.35) (26.82^2) = 7108 Joules
7108 / 0.00134102209 = 5.30 HP
34x10.50R17/D 33.5” 55.1 lbs (24.99 kg)
E = (3/8) (24.99) (26.82^2) = 6741 Joules
6741 / 0.00134102209 = 5.03 HP
Based upon these numbers alone, it appears to me that the least expensive tire size larger than stock in terms of horsepower loss is the 34 x 10.50.
I realized at this point that I have officially lost my mind and need to get a hobby. Nevertheless, I am still interested in understanding the effects (in numbers) of tire size on HP and torque.
Here’s where you come in... realizing that weight is only one component leading to a loss (or gain) in HP and/or torque, how do I account for the change in tire diameter? We can deal with tread width, rolling radius, etc. later.:blah:
Come one 1Engineer, show yourself. I know you’re out there :yield:
"Suppose the moment of inertia of each wheel (rim plus tire) is I = (3/4) m R2
where the 3/4 is a coefficient related to how the mass is distributed radially, and R is the rolling radius.
The energy of one wheel rotating at a velocity of w (in radians per sec) is
E = (1/2) I w2 = (3/8) m (Rw)^2
But Rw is just the velocity of the rolling wheel with radius R, so Rw= v (meters per sec). So for one tire, the energy is E = (3/8) m v2
For a wheel (rim plus tire) mass of X pounds (X Kg) and 60 mph (26.8 meters per sec)
E = (3/8) (X) (26.82^2) = Y Joules
1 joule = 1 watt
1 watt = 0.00134102209 horse power
If we had to accelerate four wheels from 0 to 60 mph in 4 seconds, the power would be P = 4 x Y Joules/4 sec = Z watts = HP, INDEPENDENT of rolling radius!"
Reference https://www.physicsforums.com/threads/calculating-torque-hp-losses-from-wheel-choices.334984/
I used the specs for the BFG AT KO2 in several sizes and plugged them into the formula above for comparison sake. I came up with the following:
LT265/70R17/C 31.6 “ 45.7 lbs (20.73 kg)
E = (3/8) (20.73) (26.82^2) = 5596 Joules
5596 / 0.00134102209 = 4.17 HP
LT265/70R17/E 31.6” 53.4 lbs (24.22 kg)
E = (3/8) (24.22) (26.82^2) = 6533 Joules
6533 / 0.00134102209 = 4.87 HP
LT275/70R17/E 32.2” 55.4 lbs (25.13 kg)
E = (3/8) (25.13) (26.82^2) = 6779 Joules
6779 / 0.00134102209 = 5.08 HP
LT285/70R17/E 32.7” 58.1 lbs (26.35 kg)
E = (3/8) (26.35) (26.82^2) = 7108 Joules
7108 / 0.00134102209 = 5.30 HP
34x10.50R17/D 33.5” 55.1 lbs (24.99 kg)
E = (3/8) (24.99) (26.82^2) = 6741 Joules
6741 / 0.00134102209 = 5.03 HP
Based upon these numbers alone, it appears to me that the least expensive tire size larger than stock in terms of horsepower loss is the 34 x 10.50.
I realized at this point that I have officially lost my mind and need to get a hobby. Nevertheless, I am still interested in understanding the effects (in numbers) of tire size on HP and torque.
Here’s where you come in... realizing that weight is only one component leading to a loss (or gain) in HP and/or torque, how do I account for the change in tire diameter? We can deal with tread width, rolling radius, etc. later.:blah:
Come one 1Engineer, show yourself. I know you’re out there :yield:
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