Calculating the speed of a bandsaw blade -- Need help

[xo18thfa]

I just bought a 14" Rigid brand bandsaw from Home Depot. The spec sheet says that the blade speed is 2700 feet per minute. When I assembled the machine and turned it on for the first time, it seemed the blade was going a lot faster then 2700 FPM. I did some math and came up with a blade speed much much higher then 2700.

Request members of the HMEM board independently do the math to see if I am off. Here are the specs:

Motor speed, no load: 1725 RPM
Pulley diameter on the motor: 2"
Pulley diameter on the bandsaw shaft: 5.5"
Blade length: 93.5"

Many thanks in advance, Bob

 

[imagineering]

Hi Bob,
We would also need to know the Diameter of the Blade Drive wheel so you could calculate;

Motor Rpm / Pulley Ratio = Drive Wheel speed
Therefore; 1725 rpm / 2.75:1 = 627.27 rpm, (divided by, as the Ratio is geared down).

627.27 X Circumference of Driving Wheel = Blade Speed
Assuming, (for the sake of this equation), 12" diameter, (6" Radius), Drive Wheel, (which is probably wrong for this Machine)
627.27 X 37.699" (2PiR) = 23647.45Ipm (1970.6 Fpm)


This Calculation assumes no Blade slip on the Drive Wheel - The Blade length is irrelevant.

Murray

 

[AR1911]

Look at KMOT's gizmo he just bought from HF.

http://www.homemodelenginemachinist.com/index.php?topic=12164.0

Looks like what you need for this.

 

[Dan Rowe]

Bob,
Using the logic provided by Murray and using 7" for the radius of a 14" band saw I get 2299 Fpm.

Edit: I had to fix the math.

Dan

 

[xo18thfa]

Hi Bob,
We would also need to know the Diameter of the Blade Drive wheel so you could calculate;

Motor Rpm / Pulley Ratio = Drive Wheel speed
Therefore; 1725 rpm / 2.75:1 = 627.27 rpm, (divided by, as the Ratio is geared down).

627.27 X Circumference of Driving Wheel = Blade Speed
Assuming, (for the sake of this equation), 12" diameter, (6" Radius), Drive Wheel, (which is probably wrong for this Machine)
627.27 X 37.699" (2PiR) = 23647.45Ipm (1970.6 Fpm)


This Calculation assumes no Blade slip on the Drive Wheel - The Blade length is irrelevant.

Murray

Wheel dia is 14"

 

[MachineTom]

Well at 2300-2700 sfpm what you need is a speed reducer of about 1:40 to cut steel. It takes all the fun out of a great deal like that, to need to come up with a speed reducer to make it work.

 

[xo18thfa]

I came up with 2299 now too, but in a different way. The drive wheel is geared down 1:2.75. So the drive wheel is turning 627 RPM. The drive wheel is 14" diameter or 44" circumference. That circumference makes a ratio of 1:2.125 on a 93.5" blade. So the blade is turning 295 RPM. 295 RPM x 93.5" blade = 27,582.5 IPM or 2299 FPM

I minored in math in college. It was "new math", no numbers, just Greek letters.

 

[mklotz]

The length of the blade is totally irrelevant. The speed of the blade is the same as the surface speed of the blade drive wheel. (Were it not so, the blade would slip on the drive wheel.)

So we have:

SFPM = wheel circumference * RPM = (44/12) ft * 627 rpm = 2299 SFPM

The fact that your calculation uses the length of the blade proves it is wrong. Were I to replace the blade with one twice as long, the SFPM would remain unchanged yet your calculation indicates that it would double.

 

[xo18thfa]

I see what I did wrong. I included blade length, then took it back out. It was a redundant step.

Thanks.