17 replies to this topic

[Daffyflyer]

HI...
i am currently working on a Game "Sim car " (new webpage soon) it is a car tycoon car and has some deep tech detail
and i need help with the torque calculation system.
my question is this..... if i divide the torque of a real engine in NM by it's stroke in M will i get the force on the conrod/piston in N... and vica vercia?

Also any links to in depth engine/car technical articles would be great...
Cheers
Andrew

Edit: also, any idea how to calculate max port-flow from port area, port length,and port Cd?

[clSD139]

I recently bought a playstation, the response on level of chargechange is realistic on GT4. Also a high-reving enige can be experienced with the force-feedback. The steer gets lighter when you accelerate, heavier in turns in the newest versions. But transfer the edge-feeling of variable throttle and steer combined, is unseen in race-games, of simulators. Let alone complex engine dynamics. I know rollercoaster tycoon and stuff, maybe this is interesting to you:


HP=Torque x RPM / 63

http://www.bostongea...gear_theory.pdf (very end of document, right)

[david_martin]

HP=Torque x RPM / 63

You sure about that?

[clSD139]

I've never been very strong in equations, but my mentioned site prooves it! It should be very clear with CVT.

[david_martin]

The power developed in a rotating shaft is proportional to the applied torque and the rotation speed.

In derived SI units this is written as

power (Watts) = torque (Newton metres) * angular velocity (radians per second)

The conversion factor from RPM to radians per second is 2 * pi / 60 = 0.104719755 or approximately 11/105 if pi is taken as 22/7.

In Imperial units the conventional unit of power is the horse power (33000 pounds force foot per minute) and the conventional unit of torque is the pound force foot. In this case the conversion factor, therefore, becomes 2 * pi / 33000 = 0.00019039955 or approximately 1/5250 if pi is taken as 22/7. The usual conversion is taken as

power (HP) = torque (pound force foot) x RPM / 5252

[WPT]

The horsepower equation is Hp=(T*RPM)/7121 when T is in N*m.
The mean effective force on the piston for a 4-stroke is F=(4*Pi*T)/S, where S is the stroke and T is the torque of the engine divided by the number of cylinders. However, the force on the piston during the power stroke is no where close to being constant. The max force is probably an order of magnitude larger than the mean force F calculated above. The force on the conrod would require considerations of inertia forces due to piston acceleration, and some trig calculations. I suppose the air-standard otto cycle could be used to perform a paper calculation for the force on the conrod.
WPT

[clSD139]

I've seen calculations (not very complex though) and I agree this theory works.

(B)HP=Torque (Nm) x RPM / 62,025

[david_martin]

Originally posted by clSD139
I've seen calculations (not very complex though) and I agree this theory works.

(B)HP=Torque (Nm) x RPM / 62,025

1 pound force foot = 1.355818 Newton metres, which give the result WPT posted above. Would you care to explain where your conversion constant comes from and how it relates torque and power in a rotating shaft?

[Lukin]

WPT and David_Martin are right.

Power=Torque x Engine Speed
P(W)=T(Nm)xRPMx(2*pi/60) <== Converstion to rad/sec
P(kW)=T(Nm)xRPMx(2*pi*.001/60) <== Converstion to kW
P(Hp)=T(Nm)xRPMx(2*pi*.001*1.34/60) <== 1 hp = 0.746 kW

(2*pi*.001*1.34/60) = 1/7126

Therefore;
P(Hp)=T(Nm)xRPM/7126

[clSD139]

I know I suck, again I can't find anything to back it up. But a fixed relation seems reasonable. F=m x a, F is in newtonmeter and acceleration in m/s² (m/s^2?). HP is a measure unit for speeding up so there is a direct connection with momentum in the formule. I know it's more complex to prove than let's say the Kw to HP tie.

[Lukin]

I've got no idea what your on about.

1) Force is in Newtons. Torque (or a moment) is in Newton-Metres.

2) That gear theory, your confusing the terms, the equation you posted: (B)HP=Torque (Nm) x RPM / 62,025 isnt listed. Double check it.

[clSD139]

If the net force (in newton) of a rotating shaft is demanded the expression is in Nm. If the ability to go faster is demanded HP or KW is used. I don't know the Pound-unit, but it can't be that difficult.

[Lukin]

Your talking about inerta.

Newtons second law applies linearly (F=ma) or in rotation: Torque = Inertia x Angular Acceleration

This can be applied to a gearbox, engine etc. Inertia is the radial distribution of mass and (similar to mass in the linear sense) it resists the acceleration of the system. But how the hell did you get here from basic shaft power?

What are we talking about again?

[clSD139]

Well, think about an electrical engine. P (Kw)= U(v) x I(a). Now, if I want to know how much power I consume it's in Kw/h, not very difficult. You don't even need the RPM to figure out how much torque some electrical engines deliver. Just put a torque-wrench on it and read the value when the axis blocks. Alright, some engines can be destroyed, some are too powerfull. But also again, I can imagine a stable link between HP and Nm.

[david_martin]

Originally posted by clSD139
Well, think about an electrical engine. P (Kw)= U(v) x I(a). Now, if I want to know how much power I consume it's in Kw/h,

No it isn't. It is kWh - ie. power times time (by the way for dimensional consistancy you need a factor of 1000 somewhere on your RHS, but I digress). What part of this don't you understand?

• The units of work and energy are fundemental and underived, whether they are in SI, Imperial or some other units system is irrelevant.
• The units of power are always the units of energy divided by the unit of time.
• No matter what self consistant units system you choose, the units of torques and moments will eventually reduce to the unit of force times the unit of distance, which as a consequence of Newton's Second Law, itself reduces to the units of energy.

These irrefutable concepts imply that the relationship between any power and any torque will involve the dividing torque by time. The rest is simply units conversion.

But also again, I can imagine a stable link between HP and Nm.

There is, it just isn't any of the variety of relationships you have offered us in this thread. Care to propose any more?

[clSD139]

There is a spell check, maybe I need to use that every time. Take some examples of Hp Nm graphics and calculate and you'll see. The formula I mentioned seems usefull anyway. How confusing kW and HP can be! I don't know anymore why maximum Nm and HP are shown. And I'm pretty sure, a you won't feel a HP more or less!

[Fortymark]

What´s the maximum torque you can get from an N/A engine from a certain capacity?
I have seen figures up to 110Nm from 1000cc, but the highest number have been from F3 engines.
Something like 250Nm from 2 liters. Is that really possible?

Twostroke engines seems to have 1.5 X the capacity. Like an 500cc engine have about 75Nm.

[Greg Locock]

If you want maximum torque from a given engine capacity then you'd use whatever it takes to get a high bmep, which implies an efficient inlet and exhaust, which can only really be optimal for one speed. The running speed would also be fairly low, to minimise friction, since bmep=imep-fmep. imep is controlled by your combustion process.

Uisng normal techniques such an engine is not really economically viable since it would not be very powerful for its physical size, and cost. Lack of torque can be compensated for with gears, lack of power can not. That's why, particularly with small engines, they tend to be buzzy.

One of the laws of spacecraft design applies, "in practice the optimum is rarely found at the extreme bounds of a solution space" http://pages.cpsc.uc.../AkinsLaws.html

Your two stroke has twice as many firing strokes per revolution, hence, crudely would be expected to produce twice the torque. It is relatively inefficient, so it has a low bmep, so it does not achieve 200% performance.