Calculating HP

Brownout

Joined Jan 10, 2012
2,390
Ah, so torque * motion = work. Got it.
The fundamental relationship is work(W) = force(f) * distance(d). When calculating the amount of work done by a 'twisting' force (torque) we much consider the distance from the center of the force giving arise to the work done:

T = f*r, where r is the radius of the circle swept at the end of the torque 'arm'. In this case, the 'distance' quantity from the above equation for work becomes 2*pi*r (one complete revolution). So,

Work = f*2*pi*r = T*2*pi for a complete revolution.
 

shortbus

Joined Sep 30, 2009
10,045
@WBahn. This is not something that I made up. It is a well known fact that any internal combustion engine engineer, engine builder or hot rodder knows. And any dyno chart of any gasoline or diesel engine shows it. Eric gets it, sorry that you don't.
 

WBahn

Joined Mar 31, 2012
30,077
@WBahn. This is not something that I made up. It is a well known fact that any internal combustion engine engineer, engine builder or hot rodder knows. And any dyno chart of any gasoline or diesel engine shows it. Eric gets it, sorry that you don't.
Okay, so educate me. What is the meaning of these two lines crossing? Would they cross at this same rpm if you were someplace where power was measured in kilowatts and torque was measured in newton-meters? If not, would this particular rpm still have any significance whatsoever?
 

shortbus

Joined Sep 30, 2009
10,045
OK, you win. It must be very burdensome to have to be the only one with all the answers. I'm not married to you so won't argue just for the sake of arguing.
 
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Thread Starter

Dynamic_Thom

Joined Jul 16, 2015
7
Wow this kind of blew up.

I was working with our sales guy trying to work some stuff out for a customer between whether this hp motor would be better or that hp motor would be better to provide the torque we needed when he asked me where 5250 came from. Realized I was not sure so I went searching for the answer and looks like I found it.

Thanks guys,
 

WBahn

Joined Mar 31, 2012
30,077
I sure hope a metric dyno doesn't work in kilograms (a unit of mass) and N/M (which is who knows what a unit of -- force per length? assuming M is actually m).

And, again, what is the significance of this "crossing point"?
 
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