Max Torque or Max Horsepower?

[Lazy JW]

A question for all you Dyno jockeys. I'm pulling a load of hay up Mica Flats hill in third gear, wide open at 35 mph. By my calculation this is about 2280 rpm. When an engine is loaded thusly, does it stabilise at the max horsepower level or the max torque level? Or someplace in the middle? Or maybe where the two lines would cross on a graph?
Joe

 

[twentyover]

Actually, none of the above. It's stabilizing where the rate of torque required to increase speed does not exceed the torque required to increase speed.

Say that you have a torque curve that is just slightly more than flat at this speed. Another few rpm and you may have another ft-lb of torque- but those few rpm may need an additional couple of ft-lb of torque to achieve to overcome road load.

Torque will always equal HP at 5252 rpm, although usually this is beyond torque peak. HP increases (uaually) beyond torque peak as torque falloff is usually slower than rpm increase.

Have no clue if this answers your question

 

[xctasy]

Ha ha.

There is a 'rimpull' curve that Cat 789's use to work out cartage an haul times. The gradient is calculated, and then you look at the gearing, and the actual load the engine is under is calculated. This will determine exactly how much load there is on the engine. From this, the speed can be calcualted

Whats hapening is the power of the engine at those revs has been sapped by

a) areodynamic drag (very slight)

b) tire rolling resistance (considerable with the weight, size and load)

c) the drive train losses, about 27 % on a manual rwd truck

d) the air density alters the drag ( a tiny, and negligable amount)

e) the gradient (the perecntage gradient adds weight to the vehicle. A 12% gradient adds a proportianl extra weight to the vehicle, equal to some thing like Weight x Gravity x the sine of the gradient in degrees)

I'm guessing, but what you'll find is that at 2800 rpm, there is about 85 hp or so at the flywheel, or perhaps 159 lb-ft at 2800 rpm , and this is how much power is consumed by the total of those items a) to e) above. If you want 50 mph, find another 20 hp!


I'm gonna pike out there, but its very easy to calculate all this using Bowling and Gripo's website, found by clicking on to the Ford Six.Com oval and going to the tech forum, or look at this old post. http://fordsix.com/forum/viewtopic.php?t=1708 . I have updated to imperial now, and it is covered under the vortual realitiy drag strip thread http://fordsix.com/forum/viewtopic.php?t=6711.

NB// New imperial formula is

Quote:
target speed 3*drag factor*frontal area(found in books) all over 147773 to get rear wheel horsepower reguired. To that you must add required tire rolling resistance. Target amount is 8 hp for a 195 section tire on a 2000 pound car


Example for 135 mph target on a Willys drag racer with a 0.45 cd, and 25 feet of frontal area:-

(135*135*135*0.45*25)=27679219. Then this gets divided by 147773.

Rear wheel power needed is 187.3 hp, plus 11.6 hp, which is 198.9 hp.

Solve for 185.1 hp, try speed and rolling resistance values until the equation equals 185.1 hp

Doing yours, I'd say a cd of 0.70, a 30 square foot of frontal area, about 5 hp for tire loses , and that all gets factored into the equation

(35*35*35*0.70*30)=900375. Then this gets divided by 147773.

Rear wheel power needed is 6.1 hp, plus 5 hp for tire drag, which is 11.1 hp. That is about 14 hp at the flywheel. The gradient and weight combined is therefore responsible for the loss of 71 hp!


Optionally, I could do it my self, but I'm busy working out how to programe a CNC profile cutting for 64 Ecconolines X-flow transmission adaptor.

Excuses, excuses...

 

[Anonymous]

The more you load an engine the more torque (and therefore hp) you make. Good luck figuring out how much you've got at that exact moment.


-=Whittey=-

 

[Anonymous]

Lazy JW, lots of good replies here, but I'll come at it from a different perspective. The situation you describe is virtually identical to what in aeronautical engineering is solved by a 'power required' equation, and the answer is expressed in horsepower.

Horsepower is a measure of work being done over time (even very short intervals of time). When the collective drags on your engine (tires, gravity, drive train, etc) exactly offset the engine's power output, the speed stabilizes. Xecute was telling you the same thing...I just used fewer words than he did (for once...).

 

[Lazy JW]

Thanks everyone for the replies. I'm not too concerned about exact numbers, just curious about general conditions. I do know from experience that if the load increases even slightly at that point that you had better grab another gear quick because trying to start a load like that up a 7% grade from a dead stop is no fun at all. I need to find a good NP435 transmission as my T-18B doesn't have as low a gear in first. When I swapped the oil-slobbering 360 out of this truck several years ago the 1978 F-150 donor pickup had a NP435 in it. I kept the T-18 out of ignorance, thinking that Ford probably put stronger units in the 1 ton rigs. Little did I know But thanks to you folks I have picked up quite a bit of good knowledge.
Joe