BS Blade speed, help calculate!

[Roger Ronas]

I need help to calculate my BS speed.

Have 14" bandsaw with 3/4hp motor.
RPM = 1725
1.5" pulley on motor shaft
14" pulley on lower wheel
14" wheels

I come up with 450' per min.

I used to know how to cal. but that was 30 years ago.

To setup for resawing I understand you want about 3000-4000 FPM, so what size motor pulley should I get.

Show you work so I can relearn this again,

and THANK YOU

Roger

[Peter Quadarella]

So, if it spins around 1725 times per minute, and the wheels are 14" wide (diameter)... The circumference would be pi*14=~44", 44x1725=75,900 inches per minute /12 = 6325 feet per minute.

It seems like I should remember some formula for tangential velocity but it's all gone from my mind now. Hopefully the above is correct, I used to be halfway decent at that once.

[Peter Quadarella]

Oh I realized I didn't answer your question. I'm not sure what size pulley you need as I don't really understand the mechanics of the saw well enough, but if my other calculations were right you would need to slow the saw to about 950RPM to get around 3500FPM.

[Roger Ronas]

So, if it spins around 1725 times per minute, and the wheels are 14" wide (diameter)... The circumference would be pi*14=~44", 44x1725=75,900 inches per minute /12 = 6325 feet per minute.

It seems like I should remember some formula for tangential velocity but it's all gone from my mind now. Hopefully the above is correct, I used to be halfway decent at that once.

Well Peter,
At first that may look correct, but, the 1725 is only the 1.5" pulley and then goes to the 14" wheels.
I need to figure out the ratio from the 1.5 to the 14" wheels.
I don't know if you need to cal the circum or just use dia.

I will agree that it is somewhere between my cal 450 FPM and your 6300 FPM though.

Thanks

Roger

[Charlie Plesums]

If your motor pulley is 1.5 inches in diameter, the circumference is pi * 1.5 or about 4.7 inches. Each minute you will turn the pulley 1725 times, thus moving the belt about 8000 inches per minute, divided by 12 is 677 feet per minute. If the pulley on the wheel is 14 inches, same diameter as the wheel (are you sure?), then the band is also going 677 feet per minute.

If you change pulleys so that the band is going 5 times as fast, it will only have 1/5 as much force going through the wood. If you are converting a metal bandsaw (with a slow blade speed) to resaw wood, you may be okay, but if it is a wood bandsaw that is running that slow, you may not have enough power...(I have a 5 hp motor on my resawing bandsaw).

If you change the pulley size, increase the size of the motor pulley, since larger pulleys transfer power more efficiently.

[Nathan Camp]

Roger and Peter,

The motor has a 1 1/2" pulley on the shaft, and a 14 " pulley on the wheel. So each rotation of the shaft should move the big pulley 1 1/2"

If your number was right, then 6325 feet per minute would be multiplied by 1 1/2" / 14" which gives 677 fpm, with a pulley ratio of 0.107. To make the wheels spin faster than the motor, then the pulley on the motor would need to be larger than the one on the wheel. (Never mind that part, I left the diameters out of my thinking).

Remember that the motor can only do so much work (3/4 hp). With a small to large pulley ratio, you increase the torque and decrease speed. With a large to small ratio, speed increases, but the torque decreases. With a large enought ratio, you can stall a 3/4 hp motor with your hand (I would not recommend trying this at home).

Is the lower pulley really 14"? The pulley would have to be larger than the wheel to have an "id" of 14"

nathan

[Roger Ronas]

Thanks guys,
I just did another calc and came up with 677 myself.
Okay,
So if I take
1.5” x pi(3.1416) = 4.7124
14” x pi(3.1416) = 43.9824

Divide 43.9824 / 4.7124 = 9.3333

1725rpm / 9.333 = 184.82803

184.82803 rpm x 43.9824 = 8 129.18035 inches per min

8129.18035 / 12 = 677.431696 FPM

Does this look correct?


The motor pulley is 1.5" dia on the outside , and the pulley mounted on the lower wheel shaft is 14" outside dia.
Just bought this saw not sure if it's a metal or woodworking saw.
It's a Chicago Power Tools 14" supposedly from HF years gone by.

Thanks

Roger

[Alan Schaffter]

Are you sure the lower pulley is 14"- that is HUGE!!! I suspect your lower pulley is somewhere around 4". Are you sure you don't have a 3450 RPM motor too.

To calculate blade speed, first calculate the rpm of the wheel by taking into account the ratio of the pulleys:

motor speed X diameter of motor pulley divided by diam. of driven pulley = RPM of wheel.

Then convert RPM into blade FPM (RPM X circumference in inches divided by 12):

(circumference of a circle = pi X diam.)

All together your numbers look like this:

1725 * (1.5/14) X (3.14159 X 14)/12= 677 fpm

For a blade speed of 3500 FPM you would need a 3450 RPM motor, 1.5 in motor pulley, and a lower wheel pulley with a diameter of 5.4 inches.

[Nathan Camp]

Rodger,

That is mostly good, but you need the inside diameters not the outside. Assuming a 1/4" groove, that would drop the speed by ~100 fpm.

nathan

[Alan Schaffter]

Thanks guys,
I just did another calc and came up with 677 myself.
Okay,
So if I take
1.5” x pi(3.1416) = 4.7124
14” x pi(3.1416) = 43.9824

Divide 43.9824 / 4.7124 = 9.3333

1725rpm / 9.333 = 184.82803

184.82803 rpm x 43.9824 = 8 129.18035 inches per min

8129.18035 / 12 = 677.431696 FPM

Does this look correct?


The motor pulley is 1.5" dia on the outside , and the pulley mounted on the lower wheel shaft is 14" outside dia.
Just bought this saw not sure if it's a metal or woodworking saw.
It's a Chicago Power Tools 14" supposedly from HF years gone by.

Thanks

Roger

Yup you got the right numbers. Some of those Harbor Freight bandsaws are for cutting meat! (FYI, you don't need pi in the first part of the equation- they just cancel each other out!

[Greg Funk]

The circumference of the wheels is proportional to the diameter so you can just use the ratios of the diameters to determine the speed of the larger wheel. The problem is you need to measure the pitch diameter (http://www.gizmology.net/pulleysbelts.htm) which is not exactly the same as the outside diameter. To a reasonable approximation the motor pulley should be about 8.8".

Blade speed = (8.8 / 14) * 2 * pi * (7" / 12) * 1725 = 3975fpm.

Greg

[Roger Ronas]

Okay,
Pics of the saw tomorrow.
14" dia wheel and all. I'll get the motor plate as well.Thanks for all the help.

This may not have been a good buy at $10.00, if it's a meat saw but I dont think it is.

Good Night.

Roger

[Cary Swoveland]

...8129.18035 / 12 = 677.431696 FPM

Does this look correct?

Yes, but I think eight significant digits would have been plenty.

I like Charlie's explanation better. If the big pulley is the same size of the wheel (approx.), the speed of the blade depends only on the size of the smaller pulley.

Cary

[Tom Veatch]

BSFM = Blade Speed (Feet per Minute)
RPM = Motor Revolutions per Minute
D1 = Pitch Diameter of Drive Pulley
D2 = Pitch Diameter of Driven Pulley
DW = Diameter of Bandsaw Wheel (inches)
Pi = 3.14159....

Pitch diameter is smaller than the outer diameter and greater than the diameter at the bottom of the belt groove. However for this exercise, simply using the outer diameters of the pulleys is close enough. The unit dimension (ft, in, mm, etc.) of these pulleys is immaterial as long as the same unit is used for both drive and driven pulleys.

BSFM = RPM * (D1/D2) * PI * DW/12

The numeric answer is left as an exercise for the student.

[David Beverly]

Hi Guys,

You might be able to use this spreadsheet.