re: Re: more horsepower (defs)

List,

Okay, let's work the other way. What is the torque required of my
router at 25k rpm , to lift 825 lbs.(1.5 x 550) 1 foot in one second?

25000/60= 416.666666666666666666666666666667 revolutions per second

1 foot/416.666666666666666666666666666667 = 0.0024 parts of a foot x
12 (convert to inches)= .00288 circumference to 'pull' a line one
foot in one second at 25K rpm.

.00288/pi= .0009167 diameter /2 = .0004583 radius (will be used for
ft.lb. calc later).

So a 25000 rpm shaft of .0009167 diameter will 'pull' a 'line' one
foot in one second.

Lets assume 825 lbs. load on the line. It's hi-strength stuff :-)

Our load now equals 1.5 HP. Let's figure the torque.

We have 825 lbs. acting against a lever of .0004583 length. The hi-
strength 'line' has NO diameter! This is some good stuff!:-)

Since we want torque in ft. lbs. , we need to:
12/.0004583= 26179.9387 to convert the lever into one foot length. We
divide the 825 lbs. by the same factor.
825/26179.9387= .0315 ft. lbs!!! which equals 6.048 oz. in.

Hey! I've seen these digits before!

Could it be true that just over 6 oz. in., acting at a rate of 25000
rpm, equals 1.5 HP???

Something about this feels wrong. Despite the '315'. Could someone
check this math for me?

Ballendo