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I am going to use an equation $$\text{torque} = \frac{\text{power}\times 5252}{\text{RPM}}$$ derived on Wikipedia. Suppose that $\text{power} = 100\ \mathrm{hp}$ and $\text{RPM} = 5252\ \mathrm{rpm}$ $$\text{torque} = \frac{100\ \mathrm{hp} \times 5252}{5252\ \mathrm{rpm}} = 100\ \mathrm{ft\,lb}\tag{a}$$ So far, so good.

Suppose, however we first convert power from $\mathrm{hp}$ to $\mathrm{kW}$ $$1.0\ \mathrm{hp} = 0.746\ \mathrm{kW}$$ And use it as-is in our original equation with no regard to other units: $$\frac{5252 \times 74.6\ \mathrm{kW}}{5252\ \mathrm{rpm}} = 74.6\ \mathrm{\frac{kW}{rpm}}$$ So far so good (aside from out-of-place hp to kw conversion)

Now, the bad:

However, say we interpret that last number as $\mathrm{ft\,lb}$ and convert it to $\mathrm{N\,m}$ $$74.6\ \text{(interpreted as ft lbs)} = 101.1\ \mathrm{N\,m}\tag{b}$$ Using $1\ \mathrm{ft\,lb} = 1.36\ \mathrm{N\,m}$ for conversion.

Question:

Why is the result gotten in (b), $101.10$, so close to the one gotten in (a), $100$? (they are off by 1.1%) Is there a deeper meaning here or is it just a physical coincidence?

I got this example from a real world scenario where I did the above misinterpretations and conversions at the wrong time, but then got a similar looking results to where at first I thought I had rounding errors introduced by a computer. After some research I saw that this is not an a computer error. I was curious to see if I just hit on this by accident or if this closeness in results has some deeper meaning.

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Just an accident. You've discovered that the ratio of $1\text{hp}$ to $1\text{kW}$ (0.7457) is pretty close to the ratio of a $\text{N m}$ to a $\text{ft lb}$ (0.7376). So if you apply one of them and the reciprocal of the other, the answer doesn't change much.

As the value for horsepower isn't derived from other units (it's a measured quantity), there's no deeper meaning to be found.

There are many, many different ratios you can form from different conversions. To find that one particular pair falls within 1% of each other is to be expected.

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However, say we interpret that last number as ft * lb

A kilowatt hour per rpm is not a foot-pound - you cannot "interpret" it as that.

$$W=\tau\cdot\theta$$

$$P=\frac{W}{t}$$

$$P=\frac{\tau\cdot\theta}{t}=\tau\cdot\omega,$$

where $\omega$ is your angular frequency in rad/s (about 500 rad/s, in the case of 5000 RPM).

Plug in 100 horsepower for your power to find the torque is

$$\tau=\frac{100 HP}{523 rad/s}\approx0.191\textrm{ foot-pounds,}$$ which is very different than your interpretation.

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    $\begingroup$ I question your calculation of 2M rad/s. Perchance did you multiply by 60 when you should have divided? $\endgroup$ – DJohnM Sep 11 '15 at 20:16
  • $\begingroup$ @DJohnM, yes, good call. Updated. $\endgroup$ – user1717828 Sep 12 '15 at 3:42

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