I am wondering how much horsepower would be needed, or generated (not sure which term is right), if a diesel car with a 250HP motor, and weighs 3,500 pounds, is going up a slight incline (maybe 2 or 3 degrees) averaging 20 mph? Thanks.
The minimal power is given by the usual formulas for an inclined plane, but cars are quite inefficient and you have additional losses for friction, air drag, and other losses that are impossible to quantify with the parameters you gave.
A first cut would be to just estimate the work done raising the car as it goes up the incline. 20 mph is about 30 ft/sec and a 3 degree incline is about 1 foot in 15, so the car is raised 2 ft per second. You will find that the horsepower equivalent is surprisingly modest.
Thanks for the info! Is it at all possible to get a ballpark number for horsepower based on the parameters I have given. If so, a thousand thanks!
In 60 seconds, your car is going up 2ft/sec x60sec =120 feet. It weighs 3500 pounds, so the work done in one minute is 3500x120 foot pounds= 420,000 foot pounds/minute. Google one horsepower, it is 33,000 foot pounds/minute, so....
Etudiant has given you most of it, you might also consider adding an estimate of wind resistance. The equation to calculate the net force would be: Wikipedia: Drag Equation You can estimate the car's drag coefficient to be somewhere between 0.25 - 0.30 but is dependent on make/model and you will also need frontal area. You can look up some typical make/model drag area values (product of C.d and A) here: Wikipedia: Automobile Drag Coefficient Once you calculate the drag force, calculate power (Power = force * velocity); for 20 mi/hr this number may be small, but at 40 mi/hr and up it will increase dramatically (velocity squared term).