# 14" Bandsaw with 4 speed question

• posted on July 27, 2004, 8:16 am
I just picked up the HF 14" Bandsaw with 4 speeds. I am trying to figure out the spindle RPM at each of the different belt positions. The manual lists exactly what they are, but I am trying to figure out how they calculated it.
I know how to figure out the final RPM when dealing with a drive pulley and a driven pulley. I also thought I knew how to figure out the RPM of a 3 pulley setup, but I can not seem to come up with the same numbers in the manual.
Here are the numbers. (pulleys are measured at their large diameter)
Motor: 1720 RPM Motor Pulley Diam: 1.75" (this has 3 other diameters, but I am only listing the smallest) Middle Pulley Diam: 5.3" (This is where the motor belt attaches) Middle Pulley Diam: 1.89" (This is where the spindle pulley belt attaches) Spindle Pulley: 5.97"
I figure the ratio from the motor pulley to the middle pulley is 5.3 / 1.75 = 3.03 The ratio from the middle pulley to the spindle pulley is: 5.97 / 1.89 3.16
I multiply my two ratios 3.03 * 3.16 = 9.57.
Then I divide the motor RPM by the calculated ratio: 1720 / 9.57 = 180 RPM
The manual states that in this configuration, the Spindle Speed is 600 RPM.
It also states that in this configuration the Blade Speed is 625.1 FPM. Since the diameter of the bandsaw tires are 13.75", this makes sense if I use my calculated 180 RPM as the spindle speed.
180 * 13.75" * 3.14 / 12" = 647 FPM. This is close to what is stated.
So where do they get this Spindle Speed of 600 RPM?
Thanks.
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on July 27, 2004, 11:58 am
Hi Terry, I have not gone through your calculations but even so there is a problem. The outside diameter of the pulleys is not the same as the effective diameter. If you deduct the thickness of the vee belt from the major diameter of the pulley, you will be closer to the effective diameter. Calculate speed ratios based on effective diameters and the results will be closer. Dave
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on July 28, 2004, 3:01 am

be
No, the effective diameter of a Vee beelt sheeve is, for all intents and purposes, is the same as the outside diameter of the sheeve.This is according to the Machery's Handbook, 25th edition, page 2287, table 2.
Mike
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on July 27, 2004, 1:59 pm
Terry G wrote:

Maybe they have the same guys who decide what horsepower figures to use on their compressors doing the figuring.
--
Mortimer Schnerd, RN

snipped-for-privacy@XXXXcarolina.rr.com
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on July 27, 2004, 4:28 pm
Your math looks about right to me. The manual is incorrect, which is no surprise at all. Incidentally, using the outside diameter of a sheave is generally close enough unless there is an obvious difference in belt depth. Use pitch diameter if you need to be more accurate, but you don't. In fact, pitch diameter alone is not accurate either unless you dig deeper and find the effective diameter based on the angle of departure on each sheave. Just ignore that, it ain't worth anything in this case.
-- ******** Bill Pounds http://www.billpounds.com

and
listing
1.75
RPM
RPM.
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on July 27, 2004, 6:39 pm
Thanks for the replies.
I have also been reading on using a gearbox to slow down the FPM to allow for metal cutting. I don't think I will do much of it, but it would be nice to have. I'll be doing mostly wood and aluminum, and with the 3/4" wide blade I have, that shouldn't be a problem at the fastest FPM the saw can do.
I hope to find a gearbox that I can adapt to this motor, or I might just buy another motor and swap them out when necessary. It is a bit of work, but worth it for the metal cutting capability. I figured around a 200 rpm motor would do the trick. Since my overall ratio is 9.57:1, this would give me about 75 FPM which I could adjust using the different pulleys on the saw.
My only question is torque. How much final torque is typical on the cheaper metal cutting bandsaws. If I used a motor with around 30in/lbs of torque, I would multiply this by my ratio which would give me almost 300 in/lbs of torque at the spindle (this is just an example, I really have no idea of what is necessary). Any ideas of what would be a good amount of torque?
Thanks.

figure
3
attaches)
I