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.
"Dave W" <[email protected]> wrote in message
news:[email protected]...
> 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
>
>
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
Terry G wrote:
> So where do they get this Spindle Speed of 600 RPM?
Maybe they have the same guys who decide what horsepower figures to use on their
compressors doing the figuring.
--
Mortimer Schnerd, RN
[email protected]
http://www.mortimerschnerd.com
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
I have a Delta 14" combination wood/metal band saw. Using the gearing it
will reduce the blade speed way down (about 175 feet/min). I have
approprite matal cutting blades (14 teeth/in to 24 teeth/inch). However,
when I am cutting copper, brass, aluminum, or sheet metal I don't bother
changing the speed from wood cutting speed. Only for ferrous cutting (not
sheet) do I slow it down. Just don't try it with a 3 tooth/in wood blade.
And I buy the metal blades in the same size (1/2" .025) as the wood blade
so I do not have to adjust the guides every time I change.
"Terry G" <[email protected]> wrote in
news:vjxNc.172$1o.90@fed1read06:
> 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.
>
> Thanks.
>
>
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
"Terry G" <[email protected]> wrote in message
news:AboNc.156$1o.118@fed1read06...
> 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.
>
>
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.
"Pounds on Wood" <[email protected]> wrote in message
news:[email protected]...
> 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
>
>
> "Terry G" <[email protected]> wrote in message
> news:AboNc.156$1o.118@fed1read06...
> > 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.
> >
> >
>
>