Re: [CAD_CAM_EDM_DRO] Re: Re: relay help req. (horsepower definitions)

ballendo@... wrote:

> Doug H wrote:
> >Power, as you correctly stated, is an expression of work done per
> >unit time. However, twisting equations around (for marketing
> >purposes) requires breaking the rules of mathematics that are not
> >generally considered refutable.
>
> Doug,
>
> So you're saying that my 115V,8A, 1.5hp(claimed),25k rpm Black and
> Decker model 3315 router does not compare to the motor at:
>

Not necessarily. I think many of the routers that don't bear rediculous
ratings may not be far off in power output, just that you'd better not load
them down to full power for an extended period. If you were to draw full
rated power for one full minute, either the fusebox breaker would blow,
or the motor would go up in smoke.

>
> http://www.oneida-air.com/1_5hp.htm
>
> Which shows: Motor Specs-Leeson Industrial 1.5 HP Motor 110v/220v
> 17.2/8.6 amps 1.15 service factor
>
> Because Obviously 8A does not equal 17.2A! And Watts are obtained
> from the 'PIE' equation: Power(Watts) = I(Amps) x(times) E(Volts, or
> electromotive force)

The router, I suspect, at full rated load would draw more than 8 A, and the
large Leeson AC Induction motor is not as efficient as the router motor.

You can't use these current values to compute power, as they don't specify
the phase. Both motors have some power factor, so that multiplying
amps times volts does NOT give real power consumed. The universal
motor in a router may have less phase shift than the induction motor.
Many of these single-phase induction motors have really AWFUL power
factors, so the amp requirement is VERY high, but the real power
consumed is a LOT less. Capacitor-run motors are especially
noted for this.

Jon