At 03:11 AM 8/27/2009, Rolf Lockwood wrote...
>"How much torque is involved there? That’s 
>expressed as 33,000 lb ft. We get that by 
>multiplying 330 lb (the amount the good horse 
>can move in a minute) by 100 ft (the distance he 
>can move it). Put another way, one horsepower is 
>the ability to do 33,000 lb ft of work in one minute."

You've confused force and work. It's a common 
mistake when using Imperial (foot, pound) system. 
That's because "pound" can mean either mass or 
force. Work is expressed as ft-lb, while force is 
lbs (or lbs-ft for torque).

The metric (KGS) system makes the difference more 
clear. Mass is kilograms, force is Newtons.

>He can't manage that hill at 1800 rpm where peak 
>horsepower is built because rpm and road speed 
>will fall away quickly. Horsepower just can't do 
>the work and torque is at its low point up 
>there. But once the engine falls to about 1500 
>rpm and lower, he's into torque and he'll feel 
>it pick up the load and move him to the crest.

By slowing down, he's using less power. With a 
typical diesel engine, the torque peak is down 
low, so when slowing down the need for power 
falls faster than the drop in engine horsepower. 
Remember the relationship between power, torque and RPM: P = T x RPM.

If you look at the DD15 power curves ( 
http://www.detroitdiesel.com/engines/dd15/performance.aspx 
), a drop in engine speed from 1800 to 1350 
results in a decrease in HP from 455 to 400. That's a
12% drop in HP. But, the amount of work being 
done drops by ( 1 - (1350/1800) ) 25%, because 
you're going slower (power is the rate at which 
work is done). The actual decrease in power need 
is actually more than that, because aerodynamic drag drops even faster.

It's not torque which is doing the work, but 
rather the fact that the engine has a wide 
powerband because down low, torque is the major 
contributor, while up high, RPM is.