- 10,350
- 92
- 48
- Location
- Meadows of Dan, Virginia
In my MVM article on engine braking and downhill driving (#108, April 2005), I showed how to calculate the approximate engine HP based on the grade, speed and vehicle weight.
I have often played with those numbers after recording my own driving experiences and feel that the formula is fairly accurate.
This is how it works using my latest drive up the I-77 NC to VA grade, which is 5%:
The wheel rpm was 33 MPH x 484 revs/mile divided by 60 = 266 rpm (1100-20 tires)
The force "along the hill" was 29,000 lb x 5% = 1,450 lb
The amount of torque needed to overcome this force is 1,450 x the "lever" (loaded radius of the wheel) = 1,450 x 1.75 ft = 2,537 ft-lb
Horsepower = (2,537 x 266) divided by 5,252 = 128 HP
Adding about 10% drivetrain losses would bring the needed HP output from the engine to about 141 HP.
Ryan (rmgill) provided some numbers from his run up the same mountain:
Wheel rpm = 215 (25 MPH with 900-20 tires)
Force "along the grade" = 26,000 x 5% = 1,300 lb
Torque = 1,300 x 1.58 (900-20 radius) = 2,054 ft-lb
Horsepower = 2,054 x 215 divided by 5,252 = 84 HP
Adding 10% for losses yields 92 HP.
I know I was holding back a little during my drive, the boost was about 8 psi and the EGT was about 1,100°F. Apparently Ryan was holding back too since his HP output seems a little low.
BTW, I didn't have any biodiesel left coming home and the uphill drive was done on regular diesel.
Bottom line is that if you want to figure out how much HP your engine is putting out, find a hill with a known grade and keep track of the numbers.
I have often played with those numbers after recording my own driving experiences and feel that the formula is fairly accurate.
This is how it works using my latest drive up the I-77 NC to VA grade, which is 5%:
The wheel rpm was 33 MPH x 484 revs/mile divided by 60 = 266 rpm (1100-20 tires)
The force "along the hill" was 29,000 lb x 5% = 1,450 lb
The amount of torque needed to overcome this force is 1,450 x the "lever" (loaded radius of the wheel) = 1,450 x 1.75 ft = 2,537 ft-lb
Horsepower = (2,537 x 266) divided by 5,252 = 128 HP
Adding about 10% drivetrain losses would bring the needed HP output from the engine to about 141 HP.
Ryan (rmgill) provided some numbers from his run up the same mountain:
Wheel rpm = 215 (25 MPH with 900-20 tires)
Force "along the grade" = 26,000 x 5% = 1,300 lb
Torque = 1,300 x 1.58 (900-20 radius) = 2,054 ft-lb
Horsepower = 2,054 x 215 divided by 5,252 = 84 HP
Adding 10% for losses yields 92 HP.
I know I was holding back a little during my drive, the boost was about 8 psi and the EGT was about 1,100°F. Apparently Ryan was holding back too since his HP output seems a little low.
BTW, I didn't have any biodiesel left coming home and the uphill drive was done on regular diesel.
Bottom line is that if you want to figure out how much HP your engine is putting out, find a hill with a known grade and keep track of the numbers.