blade speed/spindle speed

[wallace]

Hi all, I have just finished restoreing my sagar sawbench and I wanted to check my maths for the spindle and tooth speed calculations.

My motor runs at 2900rpm
The pulley on the motor is 5.5"
The pulley on the spindle is 3.5"
5.5 divide by 3.5 = 1.57
1.57 multiplied by 2900 = 4553rpm (spindle speed)

Diameter of blade 18" multiplied by 3.14 = 56.52" circumferance multiplied by 4553rpm = 257335.56 inches /min divide by 12 = 21444 feet per minute

I understand the maximum is 18000 SFPM surface feet per minute. If my maths is correct I'd better change the pulley
any advice would be great.
Mark

 

[Digit]

Nothing wrong with your maths Mark.

Roy.

 

[wallace]

Thanks Roy, just thought I'd better check just incase my blade imploded or something.
Mark

 

[Digit]

If it will fit use a larger driven pulley, small diameters need more belt tension due to reduced contact area with the belt.

Roy.

 

[wallace]

Unfortunately I dont think I can increase the top pulley. If I go down to 3.5" that should give me a decent speed.
Mark

 

[awkwood]

I seem to remember 12,000 ft per minute as been a good speed for general use.
It works well on my larger blades.

 

[wallace]

If I have both pulleys 3.5" it comes out at a speed of 13659, that should do nicely. And I shouldn't get too much slippage.

 

[Digit]

Sounds good, that will also give maximum belt to pulley contact. Best of luck.

Roy.