DavidTD said:
Give each engine 2500lbs to carry, make it accelerate from 1000rpm to 60 mph with a 1to1 ratio gear. The engine with the most TQ wins doesn't it?
Yes, but given equal RPM, the engine with the most torque ALSO has the most horsepower.
So your statement doesn't prove or disprove either side of the HP vs TQ war.
The posted article (thanks, Scott) is absolutely 100% correct.
The statement that is attributed to Shelby is a misnomer. It creates this false idea that somehow tq and hp are different approaches. They aren't. You can't have HP without TQ, but you CAN have tq without hp (i. e. transbrake/brakeloading launch).
People describe a "torquey" engine as having good pull in the lower RPM range, but it would be JUST as accurate to say that the engine had good low-rpm horsepower.
Moreover, someone might describe a high-winding rice burner as having all high-rpm "power", but no torque-- when in reality, they are describing the high-rpm torque as well. This is because HP and TQ are essentially the same because there relationship is constant-- HP= (TQ*rpm)/5252.
Notice that in the equation there are only two variables-- all else is constant. Thus, any change in one will produce a change in the other and vice versa.
Acceleration for a constant mass will be a function of the "area under the curve" if you plotted the HP/TQ curve against RPM. If you did this, you would note that there right off the top an advantage given to an engine with a broader RPM range where it can make tq/hp. Another way to look at it is that a CTD has an inherent handicap due to it's narrow effective RPM range (the RPM range where it can make tq/hp)
Let me continue this dead horse by illustrating that last statement. Suppose you have two engines that make a CONSTANT 200 lb-ft of torque through their operating ranges. Suppose one is a gas engine that operates from 1000rpm to 6000rpm. Suppose the other is a diesel that operates from 1000rpm to 3000 rpm. Right away, we know that the gas engine will have more HP because it's higher RPM (given constant tq).
But how would this play out in a race, given the engines are installed in identical vehicles? Well, the race would be a dead heat until both engines hit 3000 rpm. At this point, it's over for the diesel. It has to shift to the next gear (let's say it's . 5 of the ratio before it- a wide split that's OD). Since the diesel had to shift, it's now operating at exactly HALF the RPM of the gas engine.
Because both of these engines make a constant 200 lb-ft, we can see that the formula says that the diesel spinning HALF the rpm is making HALF the HP!!
Oddly enough, it's also accelerating HALF as fast as it was before. Since F=MA, A=M/F. Mass is constant, so it's all about the force. Since the diesel had to shift into a . 5 OD ratio, the Force it applies to the wheels is also HALF.
Remember, with gears, power must be constant. That means that any change in speed must produce a change in force equal and opposite. Thus, for the gears to allow the shaft to travel twice as fast, it must cut in HALF the amount of force (tq) it can apply to that shaft.
The linked article explains all this.
Anecdotal evidence backs all this up. Why is it that all racing engines are high-revving relative to a road-going engine? Because RPM helps to "make" hp, and HP is all that matters.
JLH
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