Math + Wood = Hard Question

CHScrew

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The technical R&D teacher at my school asked this for an extra credit question on the Final.

How fast is the edge of a router bit moving?

Given a blade of diameter D inches, revolving at a rate of R rpm, a point on the perimeter of the blade will travel the length of the circumference C each revolution.


Speed = distance / time
C = pi x D
pi ~= 3.1416

Speed:

Distance/revolution x revolutions/time = C x R inches/minute

What is the speed of a 1" router blade, moving at 20,000 rpm at the perimeter. Answer must be in miles per hour.
 

Chris15

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So a bit of 25.4mm diameter is moving at 20 000 RPM.
20 000 RPM = 333.333 revs / sec.
angular velocity = 666.666 π radians / sec
Velocity of a point on the bit:
v = ωd (where v is velocity, ω is angular velocity and d is diameter)
v = 666.666π * 0.0254
v = 53.20 m/s
v = 191.5 km/h

No, I do not know the miles per hour. We use the metric system for a reason:twisted:
(But I think it is roughly 120 mph)

This did not seem like the hard question it was billed as (but maybe that is just 4 unit maths being useful), was there some kind of trick I missed?
 

tenor_singer

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So a bit of 25.4mm diameter is moving at 20 000 RPM.
20 000 RPM = 333.333 revs / sec.
angular velocity = 666.666 π radians / sec
Velocity of a point on the bit:
v = ωd (where v is velocity, ω is angular velocity and d is diameter)
v = 666.666π * 0.0254
v = 53.20 m/s
v = 191.5 km/h
No, I do not know the miles per hour. We use the metric system for a reason:twisted:
(But I think it is roughly 120 mph)
This did not seem like the hard question it was billed as (but maybe that is just 4 unit maths being useful), was there some kind of trick I missed?
Just multiply the kph by .6 to get a fair approximation of mph.
 

Chris15

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Just multiply the kph by .6 to get a fair approximation of mph.
I don't do miles per hour on principle:mrgreen:. But the conversion is roughly 5/8 x km/h. The only reason I remember is that 8 km/h is a common speed limit in carparks and what not and it is a relic from the days of imperial measures where it was 5 mph...
 

tenor_singer

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I do the same thing but with track. The 1600 m event used to be the "mile", or 1.6 km was the mile. Divide by 1.6 and you'll get 1 km = .6 mile.
 

CHScrew

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north of Pittsburgh, PA
Here's the answer. I didn't get it right either.


Distance/revolution x revolutions/time = C x R inches/minute
= pi x D x R inches/minute
= 3.1416 x D x R inches / minute.

Now... in mph.

(3.1416 x D x R in/min) x (60 min/hr) = 3.1416 x 60 x D x R in/hr
= (188.5 x D x R in/hr) x (1 ft/ 12in) x (1 mi / 5280 ft) = 0.003 x D x R mi/hr.

So, a 1 inch bit at 20,000 rpm is going:

(0.003 x 1 x 20,000) = 60 mph. at the perimeter.
 

tenor_singer

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Chris's method would be a correct solving method.

He got an answer of 191.5 kph which is 120 mph.

The problem is that linear (tangental) velocity is equal to angular velocity times the radius (not the diameter), so the 191.5 (which was figured out using the formula v = wd... don't know how to get greek letters...) was two times bigger than it should have been because diameter is twice the size of the radius. Take half of the 120 mph and there's your 60.
 

Chris15

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Oops, that would be the mistake that stuffed the answer up. But based on the way they mark maths down here, I could have gotten full marks -1 for that...


And I got the Greek letters by using the Windows character map...