Consider a 24000 lb truck and trailer on a 12% grade. Using trigonometry you can easily calculate that to hold any speed on that hill a force of 2859 lbs needs to be generated at the point at which the rubber meets the road. Say the tires have a diameter of 32". The torque on the rear axle will be 3813 lb ft of torque. If we neglect wind resistance it will not matter how fast the truck is moving; if it is at a constant velocity the required torque on the rear axle will be 3813 lb ft. Different combinations of gears can be used to bring that required torque to the axle so say you have a 400 HP electric motor that makes its power at 1750 rpm. We can calculate that 400 HP is capable of producing 3813 lb ft of torque at 551 rpm. The required gear ratio to use the 400 HP is 1750/551 or 3.18:1. If a gasoline engine that makes 400 hp at 5200 rpm is coupled to that axle it also would be capable of turning the axle 551 rpm while producing the necessary 3813 lb ft of twist. A gear ratio of 5200/551 or 9.44:1 would be required to make use of that power. If a 400 hp diesel engine which makes its max power at 2900 rpm were coupled to that axle it also would be capable of producing the 3813 lb ft of torque while turning the axle 551 rpm. The necessary gear ratio to make use of the diesel's max power would be 5.26:1. All of these engines would pull the truck and trailer up this grade at 52 mph. If each of these engines have a maximum HP of 400 and are coupled to the appropriate gear boxes it makes no difference what their torque rating is; they can each pull the load up the hill at 52 mph.
