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Califbill

"Spooker" wrote in message
...

In article ,
says...

On Fri, 4 Feb 2011 09:28:01 -0500, Spooker wrote:

In article ,
says...

On Thu, 03 Feb 2011 13:42:52 -0500, John H
wrote:

I'm looking at electrical chain saws, a Poulan and a Husqvarna.

Brand Poulan Husqvarna
Item Number 118147 328965
Bar Length (Inches) 16.0 14.0
Amps (Amps) 13.5 13.5 Horsepower (HP)
3.5 9.7

They are both 13.5 Amp saws, but the Husqvarna gets almost three times
the
horsepower of the Poulan. How is this possible?

Also, the Poulan costs about $75, the Husqvarna about $270.

Manufacturers of electric tools play a lot of games with horsepower.
It's really very simple however.

watts = volts x amps = 120 x 15 = 1800 watts from a typical 15 amp
circuit

1000 watts = 1 kw = 1.34 horsepower at perfect (100%) efficiency.

Therefore the absolute maximum horsepower you can get on a 15 amp
circuit is 1.8 x 1.34 = 2.41

In reality you'd be lucky to get 2 hp at normal efficiencies.

Gearing has nothing to do with it, hp stays the same, RPM and torque
change.

Depends where the horsepower is measured.

In the case of a motor it is measured at the output shaft.
One horsepower is supposed to be 550 ft/pounds per second.

Manufactures really play fast and loose with this. They use words like
"instantaneous peak" and other meaningless terms to inflate the real
number. Power still equals force over time and "instantaneous" implies
zero time. We all know what strange things happen when you divide by
zero. That is the basis for that 1=2 formula you saw in Algebra 1.

Yep, but as to the comment about where the horsepower is measured
doesn't necessarily mean at the output shaft of the motor itself. There
is a gearbox involved and that horsepower may be a lot different. Take a
look at an outboard motor's horsepower of the engine versus horsepower
at the prop.

Is measured at the prop in outboards now, but use to be at the head. Prop
will always have less HP than at the head. Friction losses.

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