Chainsaw Tooth Velocity

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Originally Posted By: Linctex
Originally Posted By: jhellwig


Edit 3 I didn't do it right at all. 3377 ft per second.


3,300 feet per second is way faster than the speed of sound

that's high velocity rifle bullet speed


Eh. I forgot to devide by 60. The first time for some dumb reason I calculated 1.5 like it was a quarter of the circumference.
 
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Why such an easy question? A more accurate answer would require knowing the chain pitch (per link) and the number of teeth on the sprocket. Or you'd have to define sprocket effective radius in a particular way.
 
Originally Posted By: Donald
The teeth will travel far enough in one second to rip your skin apart big time.


This sprocket speed was my old Homelight Chainsaw's speed.

Some chainsaws have a rotational sprocket velocity of 7000 rpm.
 
Originally Posted By: CR94
Why such an easy question? A more accurate answer would require knowing the chain pitch (per link) and the number of teeth on the sprocket. Or you'd have to define sprocket effective radius in a particular way.


Apparently it wasn't that easy for everyone.


Not sure I understand your comment.

The sprocket rotates such that its Tangential Velocity is the equal to the chain's linear velocity, which is the same velocity for any tooth or link in the chain no matter where on the bar that tooth or link might be.
 
Originally Posted By: JimPghPA
4300 RPM/60 seconds per minute = 71.66 revolutions per second.

The circumference of a circle = pi X Diameter or pi X 2 X Radius

Circumference of a 3 inch diameter circle is 3 X 3.14 = 9.42 inches

71.66 revolutions X 9.42 inches circumference = 675.0372 inches per second.

Divide by 12 to get feet per second.

675.0372/12 = 56.2531 feet per second


and

Originally Posted By: JimPghPa
4300 rev/min x 2*pi radians/rev x 1 min/60 sec x 1.5 in x 1 ft/12 in = 56.28 ft/sec, or about 38 mph.


Quite a few responders had done the calculations properly but for some reason I could not get a simple answer to:

How many feet will any one tooth travel in 1 second?

If the chain's linear velocity is 56.25 ft/sec then the distance a tooth will travel in one second is;
56.25 ft/sec X 1 sec = 56.25 feet.
 
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Another way of approaching the problem is:

Data: Sprocket RPM is 4300, Sprocket Radius is 1.5".


Anything touching the outer portion of the sprocket will by necessity have to attain a velocity equal to the Tangential Velocity of the sprocket.

The chain effectively "translates" the rotational motion to Linear Motion.


Vt = w*R, where Omega w is in radians/sec, R is in inches.

There are 9.55 RPM/Radian. (One radian is 57.3 degrees of rotation of a full circle, or 0.1592 of a circle).

1/0.1592 = 6.28; 6.28 is 2*Pi.

w = 4300 RPM / (9.55 RPM/Radian) = 450.3 Radians/sec.

Vt = w*R = (450.3 Radians/sec) * 1.5" = 675.5 Inches/Sec.

(675.5 Inches/sec) / 12 inches per foot = 56.28 Feet/Sec.

56.28 Feet/sec. X 1 sec. = 56.28 feet.
 
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How many pins are on the rim? That would determine the speed of the chain instead of just noting the size of the rim it comes in either 6, 7, 8, 9, 10, pin.

Also i prefer using a non tacky oil in any chainsaw.
 
Originally Posted by jakewells
How many pins are on the rim? That would determine the speed of the chain instead of just noting the size of the rim it comes in either 6, 7, 8, 9, 10, pin.

Also i prefer using a non tacky oil in any chainsaw.



It doesn't matter, as the linear velocity of the chain is a function of the Tangential Velocity of the sprocket. Put a red paint dot on any one tooth and find out how far it will travel in one second. If any one tooth is moving at a linear velocity of 56.28 Feet/sec. then in 1 second it will travel. = 56.28 feet.

Quote
Data: Sprocket RPM is 4300, Sprocket Radius is 1.5".

Anything touching the outer portion of the sprocket will by necessity have to attain a velocity equal to the Tangential Velocity of the sprocket.

The chain effectively "translates" the rotational motion to Linear Motion.

Vt = w*R, where Omega w is in radians/sec, R is in inches.

There are 9.55 RPM/Radian. (One radian is 57.3 degrees of rotation of a full circle, or 0.1592 of a circle).

1/0.1592 = 6.28; 6.28 is 2*Pi.

w = 4300 RPM / (9.55 RPM/Radian) = 450.3 Radians/sec.

Vt = w*R = (450.3 Radians/sec) * 1.5" = 675.5 Inches/Sec.

(675.5 Inches/sec) / 12 inches per foot = 56.28 Feet/Sec.

56.28 Feet/sec. X 1 sec. = 56.28 feet.
 
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Originally Posted by MolaKule
jakewells said:
How many pins are on the rim? That would determine the speed of the chain instead of just noting the size of the rim it comes in either 6, 7, 8, 9, 10, pin.

It doesn't matter, as the linear velocity of the chain is a function of the Tangential Velocity of the sprocket. ...
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It does matter, because the most relevant tangential velocity (that equals the chain speed) is that at the effective pitch radius of the sprocket, not at its overall outer radius.


Small difference, perhaps, but significant.
 
I think the difference is going to be very small:

For a sprocket (from the rim toward the center of the bore) you have an Outside (or Outer) Diameter, a Pitch Diameter, and a Bottom Diameter.

See Page 3 of:
Martin Sprocket Engineering

You made me open up my old HomeLite Timberman ChainSaw but hey, it needed cleaning anyway. I put the sprocket measurements and calculations into MatLab and here are the Results:

Here are the measurements of the ChainSaw drive sprocket:

OD = 1.25"
BD = 0.75"
PD = 1.125"
Sprocket RPM = 9000

Linear Velocity using
OD = 1.25"
Sprocket RPM = 9000

49.1 ft./sec = 49.1 feet linear travel in 1 second;


Linear Velocity using
PD = 1.125"
Sprocket RPM = 9000

44.2 ft./ sec. = 44.2 feet linear travel in 1 sec.

A difference of 10%.


You still need those safety Chaps and HD Gloves!
cool.gif


BTW, I need to sharpen the chain on this puppy as well.
23.gif
 
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Originally Posted by ironman_gq
That's a pretty slow saw! Most will turn 12-13,000rpm no load, some go to 15,000rpm.


True, as this is an older saw but still works well.

The main focus of this QOTD was to review basic mechanics and specifically, the conversion of Rotational velocities to Linear velocities.

Addendum: for a chainsaw with a 20" bar and 14,000 RPM in the calcs, here is the result:

Linear Velocity using
PD = 1.125"
Sprocket RPM = 14000
69 ft./ sec. = 69 linear feet travel in 1 sec.

For a 20" bar, the total distance one cutting tooth would travel from position A back to position A in the track is:

2X22" + 2" nose = 46" for a complete travel from position A back to position A.

(69 feet/sec)/3.8 feet = ~ 18 complete travels around the bar for any one tooth in 1 second. Ouch!
shocked.gif
 
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Originally Posted by Exhaustgases
Did it quick and got 56 something feet.


Which diameter and RPM did you use?
 
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I did it right then forgot to divide by 60 oops. You got feet traveled per minute.
 
Originally Posted by Cujet
I've got a radar gun, I may just try to read the speed on my chainsaw's chain.. Could be interesting.


If you are serious, be careful.

Technically, I think the radar beam will be too wide to pick up individual teeth. I.e., I doubt there will be enough resolution.

Since we're drifting here we'll lock this thread.
 
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