Dynamometer readings & 5,250 RPM

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On a dyno chart, why does the HP/Torque curve always intersect at 5,250 RPM?
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Because (Torque X Engine RPM) / 5252 = Horsepower

EDIT: I see I was late on the draw.
 
Correct, when measuring torque in ft*lbs and power in horsepower, the unit conversion causes the two curves to intersect at 5250 rpm.
 
Originally Posted By: Steve S
Who is going to cry when the price is $4.00 per gallon gasoline.


Not the people getting 25+ MPG highway with these. We easily get/surpass 28mpg highway (24 rated) with our 07 V6 Mustang auto, at 75 mph. I've seen/talked to a few people with newer GTs who can get upwards of 27 on the highway.
 
Originally Posted By: rudolphna
Originally Posted By: Steve S
Who is going to cry when the price is $4.00 per gallon gasoline.


Not the people getting 25+ MPG highway with these. We easily get/surpass 28mpg highway (24 rated) with our 07 V6 Mustang auto, at 75 mph. I've seen/talked to a few people with newer GTs who can get upwards of 27 on the highway.
Might have to put one on my cars to buy list.
 
Yep, because the units are arbitrary, HP and torque have to be the same number (not units) at some point.
 
But I was wondering why it happens at 5,250, and, got this from asking the same question at another site:

Quote:
Awhile back, a gentleman by the name of Watt (the same gent who did all that neat stuff with steam engines) made some observations, and concluded that the average horse of the time could lift a 550 pound weight one foot in one second, thereby performing work at the rate of 550 foot pounds per second, or 33,000 foot pounds per minute, for an eight hour shift, more or less. He then published those observations, and stated that 33,000 foot pounds per minute of work was equivalent to the power of one horse, or, one horsepower.

For purposes of this discussion, we need to measure units of force from rotating objects such as crankshafts, so we'll use terms which define a *twisting* force, such as foot pounds of torque. A foot pound of torque is the twisting force necessary to support a one pound weight on a weightless horizontal bar, one foot from the fulcrum.

Now, it's important to understand that nobody on the planet ever actually measures horsepower from a running engine. What we actually measure (on a dynomometer) is torque, expressed in foot pounds (in the U.S.), and then we *calculate* actual horsepower by converting the twisting force of torque into the work units of horsepower.

Visualize that one pound weight we mentioned, one foot from the fulcrum on its weightless bar. If we rotate that weight for one full revolution against a one pound resistance, we have moved it a total of 6.2832 feet (Pi * a two foot circle), and, incidently, we have done 6.2832 foot pounds of work.

OK. Remember Watt? He said that 33,000 foot pounds of work per minute was equivalent to one horsepower. If we divide the 6.2832 foot pounds of work we've done per revolution of that weight into 33,000 foot pounds, we come up with the fact that one foot pound of torque at 5252 rpm is equal to 33,000 foot pounds per minute of work, and is the equivalent of one horsepower. If we only move that weight at the rate of 2626 rpm, it's the equivalent of 1/2 horsepower (16,500 foot pounds per minute), and so on. Therefore, the following formula applies for calculating horsepower from a torque measurement:
Torque * RPM
Horsepower = ------------
5252

This is not a debatable item. It's the way it's done. Period.
 
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