## Horsepower vs Torque vs Revs

All that has to do with the power train, gearbox, clutch, fuels and lubricants, etc. Generally the mechanical side of Formula One.
Odin did you mean HP=torque*RPM/63000 or HP=(torque*RPM)/63000? I was also just wondering if you know where the figure 63000 comes from?
ib141

Joined: 10 May 2005

In the current F1 engines the CR is in the order of about 13, limited mainly for geometrical reasons.

I stand corrected!! Thanks reca.
bcsolutions

Joined: 22 Mar 2005
Location: Lincoln, UK

ib141 wrote:Odin did you mean HP=torque*RPM/63000 or HP=(torque*RPM)/63000?

actually the two options are the same thing...

The constant depends by the fact that he’s using Imperial units.

The original formula, from physics, is :

Power = torque * rotational speed

If you use SI units, torque is in Nm and rotational speed in rad/s, then you get power in Watt, no need for other constants.

To have power in hp you have to divide the result by 745.7.
Dividing it by 735.5 you get the cv, Italian abbreviation for cavalli vapore, or PS, German abbreviation for... I don’t speak German so I don’t know..., just, be careful with BMW data for power... 900 PS are 888 hp

If you use torque in Nm and rotational speed in rpm then the formula becomes :

Power = torque [Nm] * rpm * 2 * pi / 60

Again you’ll have power in W so you need to divide it by 745.7

If you use Imperial units then

1 lb is 0.4536 kg, 1 ft = 0.3048 m, 1 ft = 12 in.

1 [Nm] = 1 [lb ft] * 0.4536 [kg/lb] * 0.3048 [m/ft] * 9.81 [m/s2]

So the formula becomes :

Power = torque [lb ft] * 0.4536 [kg/lb] * 0.3048 [m/ft] * 9.81 [m/s2] * rpm * 2 * pi / 60 = torque [lbft] * rpm / 7.04

That gives power in Watt, divide it by 745.7 and you have hp so the formula is :

Power [hp] = torque [lb ft] * rpm / 5250

Since 1 [lb ft ] = 12 [in lb]

Power [hp] = torque [in lb] * rpm / 63000

Needless to say which units are more clever...

bcsolutions wrote:I stand corrected!! Thanks reca.

You're welcome.
Reca

Joined: 21 Dec 2003
Location: Monza, Italy

The original question was put forward as a rough generalization, so that is how the subject will be responded to.
TDC is top dead center, the position where the piston is at the very top of it's travel, and maximum compression occurs. At 90 degrees past TDC, the crankshaft is at right angles to the centerline of the cylinder bore.
Given a fixed displacement ( presently 300cc per cylinder, ref. 5.1.2 FIA technical regulations), then generally, if you have a stroke increase, bore will be reduced to meet the regulations, and vice versa. Increasing revolutions almost always leads to an increase in power. But as revolutions increase, it is very desirable to decrease the stroke. So that is why you will see short stroke, large bore engines used in reciprocating engines putting out as much power as possible.
Since the distance from the centerline of the crankshaft to rod journal centerline decreases with reduced stroke, it's just simple, basic leverage rules that apply. The piston is pushing down at a certain force, and decreasing the lever arm will lead to reduced torque at the crankshaft.
No, the old “for a given displacement short stroke means less leverage hence less torque” is a false myth, for a given displacement the amount of peak torque is pretty much independent by the bore/stroke ratio.

Obviously this person knows very little about the practical side of life.
Guest

Guest wrote:Since the distance from the centerline of the crankshaft to rod journal centerline decreases with reduced stroke, it's just simple, basic leverage rules that apply. The piston is pushing down at a certain force, and decreasing the lever arm will lead to reduced torque at the crankshaft.

Does the “certain force” remain the same when you reduce the stroke, for a given displacement ?
Reca

Joined: 21 Dec 2003
Location: Monza, Italy

Clear. We have to use a simple, basic, law of physics (torque = force x leverage) because it fits with your “generalization” and we have to totally neglect another simple, basic, law of physics (force = pressure x area) because it doesn’t.
Reca

Joined: 21 Dec 2003
Location: Monza, Italy

What then will the F1 engineers have to consider when designing the combustion chamber for next years V8's, as opposed to this year? Does anyone envisage any specific differences or are they effectively going to be identical?
bcsolutions

Joined: 22 Mar 2005
Location: Lincoln, UK

bcsolutions-

Engine designers have lots of things to consider when designing their race engines:

-Big bores allow for large valves (good).
-Short strokes allow for high rev's (good).
-Unfortunately, a big bore and high compression ratio leads to a short, wide combustion chamber. Such a combustion chamber typically has sluggish combustion characteristics. (Not good for a high rpm race engine).
-All F1 engines are structural members of the chassis. Thus a short, torsionally stiff V8 is better than a longer, less stiff V10.
-A 20,000 rpm 2.4L V8 producing 900 hp will have higher fuel consumption than a 18,000 rpm 3.0L V10 producing 900 hp.
riff_raff

Joined: 24 Dec 2004

This is always an interestimng subject as it seems to be a few miscinceptions around it. First of all, as stated above, Power (kW) equals Torque (Nm) times Angular velocity (rad/s).

As Mauro Forghieri once explained, "Power comes from speed, torque without speed is nothing", when he tried to explain why he went with a V12 rather than a V10 or V8 configuration for Lamborghini's 3500cc F1 engine back in 1989.
Reason was that a V12, with its lesser individual reciprocating masses such as pistons and con rods, could master a higher Rpm.

Indeed, with my bodyweight and a sizeable wrench, I could easily create more torque around a stubborn hex-bolt than any F1 engine would muster, but it would still not be any power produced as there is zero speed involved.
"Bernoulli is a nine-letter name"
xpensive

Joined: 22 Nov 2008

Bottom line, torque, power output, and gearing are all equally important. Be nuts to say any one is more important than the other.
Grip is a four letter word.

2 is the new #1.
Jersey Tom

Joined: 29 May 2006
Location: Huntersville, NC

Old indeed, as old as the invention of the Otto-motor itself actually, still interesting though.

Gearing is obvioulsy most important in order to optimize the use of the engines peak power-band, why in a perfect world with no frictional losses, a CVT should be superior to a step-by-step gear-box.
"Bernoulli is a nine-letter name"
xpensive

Joined: 22 Nov 2008

Some basic, generally accepted first principles:

An ICE is a form of air pump - the more air it pumps in and out, the more power produced. More RPM = more air pumped = more power.

A supercharger or turbocharger forces more air into the engine and creates more power.

Of course the intake and exhaust systems are, as is the head design -- all of which contribute to air flow.

Short stroke = more RPM = more HP (but less torque)
Long stroke = more torque (but at the cost of lower revs and less HP
Enzo Ferrari was a great man. But he was not a good man. -- Phil Hill
donskar

Joined: 3 Feb 2007
Location: Texas, USA

Expanding the subject now, are we, as squeezing more oxygen molecules into the combustion chamber has less to do with the relation between Tprque & Rpm me thinks?
"Bernoulli is a nine-letter name"
xpensive

Joined: 22 Nov 2008

Power = torque * rotational speed

this is why it would be so simple to limit all 2009 engines to the same power output. you just have to have a torque sensor. rpm is already measured. and all the sudden teams would fight for every ml less of fuel consumption. Isn't that what the FiA wanted. and it would be perfectly fair for all.
Formula One's fundamental ethos is about success coming to those with the most ingenious engineering and best .............................. organization, not to those with the biggest budget. (Dave Richards)
WhiteBlue

Joined: 14 Apr 2008
Location: WhiteBlue Country

Simply an excellent idea, hats off white and Blue!
"Bernoulli is a nine-letter name"
xpensive

Joined: 22 Nov 2008

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