Torque versus Horsepower

  • Thread starter Thread starter Anonymous
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I wasn't going to get into this, but...
Required flywheel weight is linked to veh weight, I'll explain;
I had a 351c/4spd in a pinto(11.21@ I think 114mph) A heavier flywheel hit the slicks harder and picked it up around a tenth.(mostly 60ft picked up)
Sold the car and put the motor in a 71 fastback mustang....LIGHTER flywheel picked it up around 3 tenths but 60ft stayed pretty close.

The heavy flywheel had more "hit" and in spite of having to accelerate the added weight, the light car was quicker.

The heavier mustang had plenty of traction, it just needed help reving, therefore the lighter flywheel helped.

You might think that a heavier car would respond better with more flywheel weight to get it moving, and a lighter car would like a light flywheel........ME TOO!

A lot of part changing proved me wrong and this was my results.

As far as the torque/horsepower argument, in general, torque engines(big) have a shorter power span in comparison to a hp engine. In other words, a 460 not only accelerates at a lower rpm per sec than a 289, a 289 will also rev from 2500-7500 where a 460 rev's from 2500-5500(for example) and will require shifting sooner.
In drag racing, a 460 will be quickest if you accelerate the engine slowly by using a loose converter and a powerglide or a higher rear gear. Otherwise you lose to much torque in spining the engine.
To prove this on a dyno, do acceleration rates of 300 rpm per sec than 700 rpm per sec.
A 289 will see less difference between the two pulls. A 460 will lose a lot of torque in the second pull because your wasting it spinning up the heavy bottom end.
In the real world, 460/c4/5:14 gear, the accel rates would be higher than the engine could produce in 1st and 2nd, so if you change to 3:91's, you would actually be quicker. A 289 on the other hand, can accel at 900 rpm per sec so it would be quicker with the 5:14's.

Anyway,enough rambling, I bracket race so I like big torque/low rpm/big cube engines over tempermental/high maint/high rpm/small blocks.
 
So do all the drag racing fraternaty. The hi-rpm power is just a crock. The old adage "For Staight line speed, go cubic inches" is the truth. A supercharger is another item used to increase power without increasing revs. A turbo can do the same. So can nitrous, although people often loose the plot here.

A 200 or 250 six fits in with the plan. As long as you maximise power without trying to make it scream too much. Heres a good idea. Aussie 265 Hemi with triple carbs, and about 250 hp net. This engine didn't have to rev to 7000 rpm to kill a 351 Cleveland or 340 Mopar. It's 60 ft time was quicker than either of the other engines. Quarter mile time was much quicker, and it was only over 100 mph that you noticed the lack of 1.5 liters and two extra cylinders! All because it was tuned to give its power by 5000 rpm. And a massive 430 lb-ft gross at a lower rpm than the others. And how does an engine behave like a tractor with a wild 302 degree cam? A six cylinder engine with good cubes, a poor rod ratio and good carbs that pulse tune well ends up lowering the peek power rpm compared to a similar sized Chevy V8.

As ever, when Ford guys follow Chrysler guys, then they start winning races. (250 followed 225, 460 followed 440, Injected 2.3 Turbo followed 2.2 Turbo)
 
I've been holding fire on this cause I've been out of town and been in a "monitor only" mode for the last few days, no time or ability to post anything lengthy.

Everybody knows by now that:

HP = T x N / 5252

Where N is engine RPM. Seems simple, the higher the RPM at which an engine delivers peak torque, the more power it makes.

There's another equation for torque:

T = (BMEP x V) / (48 x Pi)

Where T is in ft-lbs., BMEP is brake mean effective pressure and V is displacement in cu. in. This seems to say that the bore and stroke proportions of an engine are irrelevant to its torque production, all that matters is BMEP.

BMEP is the actual average pressure produced in the cylinder during the power stroke reduced by the mechanical inefficiency of the engine. It is directly related to the amount of fuel/air mixture present in the cylinder which is a function of volumetric efficiency.

VE is related to several things. First of all it is a function of intake runner, port and valve shape and size. Ricardo showed that VE was maximized at a mean intake velocity of 140 FPS and was acceptable in the range of 40-240 FPS. Velocity is a function of the volume of air flowing through a given port, so a small displacement engine is going to achieve peak VE at a higher RPM than a large one breathing through the same head. In other words a 240 will peak at higher revs than a 300, no surprise there.

VE is also a function of piston speed, decreasing as speed increases. It typically peaks at around 1600 FPM and drops below useful above 3200 FPM, although aggressive camming can alter this range appreciably. Short stroke engines operate at lower piston speeds for a given RPM than long stroke engines so again they will achieve peak VE at higher revs.

So what happens in a long stroke engine is this: It reaches peak VE and thus peak BMEP at a lower RPM than a short stroke engine. If the peak efficiencies and displacements of both engines are the same, THE PEAK TORQUE WILL BE THE SAME. Look at the nearly identical peak torque figures for the 300 and the 302. The ratio of the RPMs at which they occur is very nearly the inverse of the ratio of their strokes. Horsepower though is a function of peak torque and RPM so for two engines of identical displacements and efficiencies, the one with the shorter stroke will always make more HP.
 
I just want to be clear on this...

StrangeRanger":146yt8pp said:
... Horsepower though is a function of peak torque and RPM so for two engines of identical displacements and efficiencies, the one with the shorter stroke will always make more HP.

Regardless of how many cylinders. Correct? In other words- all else being the same, an effective means of 'de-stroking' an engine is to increase the number of cylinders?
 
If you mean, to take an example, "Would a perfectly square V-10 of a given displacement generate more power than a perfectly square V-8 of the same displacement, assuming that the porting, camming , etc was equivalent?" The answer is yes because the V-10 would necessarily have a shorter stroke and thus have its peak efficiencies at a higher RPM.

What gets interesting, to borrow an example from USAC of many years ago is that for a given displacement an angine with fewer cylinders has more room in the head for valves and can acheive higher maximum VE and thus higher BMEP and higher peak torque. This made the I4 Offy competitive with the Ford Indy V-8 because on some tracks it could exit the corners harder but could not gain speed down the straights quite as well. The Offy ruled the shorter tracks and competed well on the road courses but was toast at Indy.
 
What a bout fuel efficiency? Wouldn't an engine with fewer cylinders and a long stroke have more ability to utilize the pressure that is developed in the chamber? It seems to me that a slower turning engine will have more time for the available pressure to push on the piston. This engine will also have less surface area in the chamber relative to the volume, resulting in better thermal efficiency. An extreme example would be the huge diesel engines for ships that have the 38" bore and 96" sroke. These engines are reported to be very fuel efficient and only turn at 103 RPM. Am I all wet with my theory?
Lazy JW
 
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