Busting my brain

Hey guys and girls. I'm a second year mech eng student, working as an
assistant to engineers in the coil coating industry for the summer. We coat
aluminium and then roll form it into various building products. Naturally,
we have a great deal of high capacity coilers/uncoilers.
I wish to calculate the torque required to accelerate the shaft to the
needed 400 feet per minute feed speed, given a coil outside diameter of 55
in., inside diameter of 20 in., and a coil mass of about 5000 lbs.
What i did was just took the coil alone by itself and calculated the torque
required to accelerate it to the needed speed within 2 seconds (reality),
but i got some ridiculously huge numbers. I used:
sum of torques = (polar moment of inertia)*(angular acceleration)
Taking polar moment to be: (0.5)*
(mass)*(OD^2 + ID^2)
and angular acceleration = tangential / radius.
The kicker is we have a 14.8 lb-ft 5 hp 1765 RPM motor, through a 20:1 gear
box, then a 2:1 reduction driving the shaft. Could anyone run the numbers
for me, or let me know if i'm missing something? Maybe i just have the
blinders on..
Thanks in advance for any help,
-JP Venturi, Second year McGill engineering,
formatting link
Reply to
Loading thread data ...
Here's a non rigorous back of the envelope:
we want to move the perimeter of a coil at 400 fpm in 2 seconds Further, let us suppose that the entire mass 5000 lb is concentrated at the perimeter, then in SI f (N)= m.a = 5000/2.2 kg X 1 m/s^2 = 2309 newton Supposing the force X av speed = 2309 X 1 m/s = 2.3 kW That's a starting HP of 3.1 HP allowing as low as 3.1/ 5 X 100% efficiency = 62% in the reduction gear.
This seems within bounds, particularly as the moment is less than the linear inertia that I supposed.
Did I mention this was going to be a rough estimate?
Cross-checking now: torque available is given as 14.8 lb-ft from a 5 HP motor turning at 1765 rpm
1765 rpm = 29.4 rev/sec = 29.4 X 2pi feet/sec per foot of radius So 14.8 lb force at 182 ft/sec speed = 66 newton at 55.5 m/s = 3.7 kW = 4.9 HP Close enough!
Brian Whatcott Altus OK
Reply to
Brian Whatcott
since Brian took a crack at this so will I................
your 55 in dia (4.6') has a 14.4' circumference
so you need to turn the coil (when its largest) at about 28 rpm to generate your 400 fpm feed rate as the coil gets smaller you've got the turn it faster
so it looks like your gearing arrangement is ok when the coil is big but looks a little slow when the coil is small.
If my calcs are correct I get .467 rev/sec (2.9 radians/rec) at full speed; you give 2 seconds as the time to full speeed
this yields 1.47 rad/sec^2 angular accel
I get a rotational moment of interia of ~1420 slug ft^2
this yields a torque of 2080 ft-lbf at you coil speeds about 11 hp
looks like you don't have enough torque to do the job but if the actual performance contradicts the cacls then we need to take a closer look
one of the tricks in this situation is to not need to accel the entire coil at once. Some sort of material buffer (spring tensioned coil loop) would greatly reduce the accel (torque) demands.
cheers Bob
Reply to
Yep, Newton's 2nd law for rotational systems.
T = I*alpha
Mass moment of interia should be
I = m*(Do^2 - Di^2)/8 or m*(Ro^2 - Ri^2)/2
If you mix lb, ft, and inches in those equations a wrong answer will result. Carry units through the calculations. Cancel out units in the formula to get units for the answer.
The output shaft will have a rotational speed of 44 1/8 rpm with a torque of 59.2 ft-lb (40 times).
Reply to
Jeff Finlayson
Thanks for the replies guys, they helped out more than you can imagine. Turns out i was making some little dimensional errors. Up here in Canada we mix inches and mm's into an orgy of dimensions..you gotta be real careful
Thanks guys, much appreciated
Reply to
Brian, Jeff-
According to my calcs he doesn't have enough hp or torque to accelerate the full coil to speed in 2 seconds. Even with the gear reduction he's only got about 600 ft-lbf on the output shaft, which appears to be less than 1/3 of what is needed.
Jeff looks like your decimal point is misplaced?
cheers Bob
Reply to
a 14.8 lb-ft 5 hp 1765 RPM motor, through a 20:1 gear box, then a 2:1 reduction driving the shaft.
400 feet per minute feed speed, given a coil outside diameter of 55 in., inside diameter of 20 in., and a coil mass of about 5000 lbs.
circumference = pi D = pi * 55/12 = 14.4 ft
400 ft/min = 6.7 ft/sec
This needs a shaft rotation of 6.7/14.4 rotations/sec = 0.46 revs/sec
The motor speed of 1765 rpm is reduced to 1765/40 = 44.15 rev/min = 0.74 revs/sec so the gearing is appropriate.
Now, double check the power available. a mass of 5000lb from 0 to 400 fpm in 2 seconds takes a force of 2309 newton (518 lbf) at a speed of 6.7 / 2 ft/sec That's 1735 ft.lb/sec or 1735/550 = 3.2 HP
or in the SI manner: force of 2309 newton times av speed of 1.02 m/s = 2309 X 1.02 watts (so much simpler!) = 2358 W = 2358/746 = 3.2 HP if you insist!
What output shaft torque is that? 518 lb at radius 28 inches = 1209 lb.ft Compare this with your figure of 600 ft-lbf
Why the difference? the available torque has to be taken together with the shaft speed to arrive at HP.
On the coil acceleration, I am using constant HP, which means twice the torque at half the shaft speed!!!
Not all motors behave this way, but you are probably familiar with how domestic air compressors are rated: the so called 4 HP motor is usually a 2HP motor, and they are rating it at a short term start up load. So you can see that some motors will give increased torque at slow speed - even increased power.
I hope this helped...
Brian Whatcott Altus, OK
Reply to
Brian Whatcott
Thanks, I'll have to check my T= I * theta_dd
I came up with a required torque of ~2000ft-lbf to accel to 44 rpm in 2 secs
I must have goofed on the mass moment or theta_dd
cheers Bob
Reply to
Inertia of a coil 55 inch outside diameter 20 I.D and weight of 5628 lbs is:
705 kg-m^2 ------or-------- 16 725 lb-ft^2
using: I= (m/2)*(Ro^2+Ri^2)
This is a real coil that was run on the machine. The machine is driven by a Marathon black max inverter duty motor, which from what i understand means that it is putting out a constant torque over it's entire RPM range, up to 1765 RPM. This means horsepower is increasing linearly with RPM.
From here i calculated that i would need about 22 ft-lb of torque at the motor to yield the right acceleration. Given the actual real life gear ratio of 17.15: 1 at the gear box and 2.0:1 final reduction, yielding a total 34.1:1 reduction. About the gear box, the data plate fell off and we had to track down the real numbers, so i just made an assumption, which turned out to be pretty close.
The 22 ft-lb is about 150% of the rated value, which is within the breakaway limit of the motor. This is the starting torque, which is always higher as we have flywheel type loads. Later the torque drops to within rated values. I was later informed by our senior electrician that the motor driver is an intelligent unit, that dynamically varies the RPM of the coil with the radius to match feed speed of the machine (but remember that torque is cst over the whole range)
Other than that, i think i'm good now. Numbers are good, and i can proceed to the next phase of my little project, selecting sprockets and suitable belt drives..
Thanks guys, -JP
Reply to

Site Timeline

PolyTech Forum website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.