OK, 6 FP (foal power)
... They take motors at low temp and measure inflow current (which only lasts for about 8 milliseconds before it hits the windings). It really has nothing to do with the actual physical output of the motor itself. My vac, in reality probably outputs under 1/10th of a horse...
What you say is basically the scenario for a single phase capacitor start AC induction motor where the initial in-rush current is to provide sufficient torque to accelerate the rotor from zero to full speed in a fraction of a second. That initial current is known as the locked-rotor current because the motor does start out in a stalled or "locked rotor" condition for a couple milliseconds. But, all of this has zero to do with
OUTPUT power which is what real NEMA motor ratings are actually all about. Output power from a motor is measured by measuring both the speed and the full load torque (which both have to occur at the same time to make any sense).
Just in case anybody is curious, one horsepower is 550 ft-lbs/sec where ft-lbs are a measure of torque in the English system of units and the 1/sec term is the speed in radians per second. Unless you are an engineer, you usually think in terms of RPM rather than "natural frequency" expressed in radians per second. So, if we lump all of the various conversion factors together, we can use the following relationship:
HP = (ft-lbs X RPM) / 5252 where 5252 is the composite conversion factor.
A real world example would be one of the Baldor motors that I have where the full load speed is 1750 RPM and the full-load torque is 4.5 ft-lbs. Plugging in the numbers give us a mechanical horsepower output of 1.5 HP, which is exactly what is stamped on the nameplate.
Most of us don't have the test equipment to directly measure torque with the motor running, but we can rearrange the equation to find out what the full load torque would be if we know the rated full load horsepower and the speed of the motor under full load. My Robust lathe has a 2 HP Leeson motor with a full load speed of 1740 RPM. Suppose that I want to know what is the maximum torque at the spindle at the maximum speed of 1065 RPM in the low speed range. First we plug in the motor numbers into the equation and we get:
ft-lbs = (2 X 5252) / 1740 = 6
The transmission ratio is 1065 / 1740 = 0.612 and momentarily ignoring efficiency, the spindle torque would be:
6 / 0.612 = 9.8 ft-lbs
The drive train uses a J-section belt which has a very high efficiency of about 98%, so:
9.8 X 0.98 = 9.6 ft-lbs.
I ought to mention that this is only at full speed and only when delivering the full load torque which is hardly ever the real operating point on a woodturning lathe. On any machine with electronic variable speed (aka, inverter), the available torque below base speed (1740 RPM in this case) is essentially constant, but the first equation tells us that the available power varies in direct proportion to the speed.
So, how about the 6 FP (foal power) shop vac? The shop vac uses a universal motor connected to a small squirrel cage blower. Universal motors have some odd characteristics. Their speed is not regulated in the sense that an AC motor is. At the maximum speed, the output power and output torque are both zero, therefore it is called the "no-load" speed. The peak power is reached between 1/4 and 1/2 no-load speed and peak torque is reached at zero speed, so for obvious reasons it is called the "stall torque".
But, how much power does it take to run a squirrel cage fan anyway? In free air, not much at all, but in a shop vac there is a meaningful pressure differential of several PSI, depending on how good the particular machine happens to be. I have a very good shop vac -- a Fein 9.77.25 with a nameplate rating of 1100 watts nominal power and maximum input power of 1200 watts (120 volts @ 10 Amps). I'm proud to say that machine really sucks. I have measured something around 7 in-Hg blocked-port vacuum.
We know that the mechanical output power can't be greater than the electrical input power and the efficiency of a universal motor is all over the map from zero to approximately 50% maximum. One horsepower is 746 watts, so 1200 watts would be 1.6 HP input power. Assuming a best-case 50% efficiency for the motor of the shop vac, the maximum mechanical output power would be approximately 0.8 HP. That's not slouching, but it is shy of the 6 HP claimed by the unmentioned brand by a factor of 7.5. I am sure that the marketeers for brand X have a song-and-dance routine to waltz around that minor discrepancy.
Small pieces are a problem with a regular vacuum chuck as well. The holding power is 14lbs per square inch.
It would be a bit less than that. A hard vacuum is 14.7 PSI differential. There are an abundance of air leaks in a vacuum chucking system and practical pumps for vacuum chucking run a bit closer to 26 in-Hg (12.8 PSI).
