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Posted Mon Jun 23, 2008, 3:19pm Subject: Formula needed
Hello. Anyone here know the formula to figure out how much inch pounds of torque is needed to turn a weight on the horizontal? In this case, a cylinder, starting from a dead stop.
senorswiss Senior Member Joined: 10 Feb 2008 Posts: 109 Location: West Lafayette, Indiana, usa Expertise: I live coffee
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Posted Mon Jun 23, 2008, 4:35pm Subject: Re: Formula needed
This is something I might actually be able to help with.
The moment of inertia for a solid cylinder about its axis is: I = (1/2)*m*r^2 where m is the mass and r is the radius. If it is a cylindrical shell (hollow center), then this changes somewhat. you can find tables of moments of inertia online pretty easily.
Torque, T is equal to the moment of intertia, I, multiplied by the angular acceleration, a (alpha, but I can't type that here). That is, T = I*a.
don't confuse this with linear acceleration since I'm typing 'a' instead of alpha.
So you can find the Torque as: T = (1/2)m*r^2*a. Make sure your units are consistent here. CGS units: mass in kg, r in meters, time in seconds, a in radians per sec^2, etc.
You need to decide what angular acceleration you want in order to solve for torque. alpha(rad/s^2) is a measure of the rate of change of the angular velocity(rad/sec).
So if you want this thing to spin at 2*pi rad/s (this is one rev. per sec), and you want it to go from rest to it's maximum speed in 3 seconds, then a= (2*pi)/3 rad/s^2.
You can find alpha for any change in angular velocity (omega = w), by taking (w_final - w_initial) / time. time should be in seconds, w in radians/sec. You can convert the torque to inch*pounds in the end. I has units of kg*m^2, a has units of rad/s^2. together these give kg*m^2/s^2 which is N*m (Newton*meter), which you could plug into an online conversion application.
I'm not sure how familiar you are with these calculations, so I apologize if I'm stating things that are obvious to you.
One more thing. These calculations assume zero friction (which is why they will give zero torque needed to keep a constant angular vel). Whether or not you need to make a more robust calculation depends on your purpose, I guess...
Hope this helps. Let me know if I stated things in an unclear way.
CoffeeRoastersClub Senior Member Joined: 6 Jul 2005 Posts: 2,144 Location: Vernon Expertise: Professional
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Posted Mon Jun 23, 2008, 4:59pm Subject: Re: Formula needed
Thank you very much Jordan. I will definitely make use of the formula.
CoffeeRoastersClub Senior Member Joined: 6 Jul 2005 Posts: 2,144 Location: Vernon Expertise: Professional
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Posted Mon Jun 23, 2008, 6:02pm Subject: Re: Formula needed
Jordan,
Please explain this part of the equation in more depth if possible:
senorswiss Senior Member Joined: 10 Feb 2008 Posts: 109 Location: West Lafayette, Indiana, usa Expertise: I live coffee
Espresso: Silvia Grinder: espresso: EB, others:Maestro... Vac Pot: Moka Pot Drip: French Press, Aero Press Roaster: Modified HG/BM
Posted Mon Jun 23, 2008, 6:44pm Subject: Re: Formula needed
Len,
We can draw a strong parallel between linear and angular variables, which may help elucidate things, since people are usually more familiar with these from everyday experiences.
Linear value: variable {units} ------------------------------------ position: x {meters} velocity: v {meters/sec} acceleration: a {meters/sec^2} so acceleration is the change in velocity per unit time.
Angular value: variable {units} -------------------------------------- angular position: theta {radians} - tells you how much of one turn the cylinder has made. 2*pi radians = one revolution = 360 degrees, 4*pi radians = 2 rev, etc. angular velocity: omega {radians/second} - tells you how fast it is spinning. 2*pi rad/sec = one revolution/sec = 360 deg/sec angular acceleration: alpha {radians/sec^2} - tells you the rate of change of the angular velocity. So this dictates how quickly you are going to go from rest to your top speed.
so just as the linear acceleration tells you how quickly your linear velocity is changing, the angular acceleration tells you how quickly your angular velocity is changing.
if you wanted your angular accel to be 5 rev/sec^2, then since 1 revolution = 2*pi radians, alpha = 10*pi {rad/sec^2 }(think of it as: 10*pi rad/s PER second) So if you start with the cylinder at rest, after one second, it will be spinning at 10*pi radians/second (5 rev/s) after two seconds, it would be spinning at 2*10*pi rad/sec (10rev/s).
Let me know if I can explain further. I would be willing to take a look at your actual calculation, as well, if that helps.
Cheers, Jordan
edit: Just to be clear, in the first post, I called omega w, and I called alpha a. so written out, clearly, Torque = I*alpha, or Torque = (1/2)*m*r^2 * alpha.
CoffeeRoastersClub Senior Member Joined: 6 Jul 2005 Posts: 2,144 Location: Vernon Expertise: Professional
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Posted Tue Jun 24, 2008, 7:50pm Subject: Re: Formula needed
Weight of cylinder is 60 pounds. Radius of cylinder is 6 inches.
so: I = (1/2) * 22.22 * .144^2 which equals: I = 23.03
T = 23.03*alpha
I want the cylinder to go from rest to one rotation per second in 2 seconds, so: alpha = (2*pi)/2 rad/s^2
I am about to go further on it, but wanted to check with you to see if I am missing something in the equation.
senorswiss Senior Member Joined: 10 Feb 2008 Posts: 109 Location: West Lafayette, Indiana, usa Expertise: I live coffee
Espresso: Silvia Grinder: espresso: EB, others:Maestro... Vac Pot: Moka Pot Drip: French Press, Aero Press Roaster: Modified HG/BM
Posted Tue Jun 24, 2008, 8:09pm Subject: Re: Formula needed
Hey Len,
I don't have my calculator here, so I'm trusting google for converting. 60 pounds = 27.22 kilograms. Maybe that was just a typo in your post.
Also, for r, I get: 6 in = 0.1524 meters
Then, I get: I = 0.5 * 27.216 * 0.152^2 = 0.31438768 {kg*m^2/s^2} ---> 0.31438768 {Newton*meter}* pi{rad/s^2} = "2.782 pound inch" (I just plugged into an online torque converter instead of doing the calc, here.)
Check your numbers and see if you come into agreement with mine. If not, I'll recheck what I did.
CoffeeRoastersClub Senior Member Joined: 6 Jul 2005 Posts: 2,144 Location: Vernon Expertise: Professional
Espresso: Vintage La Pavoni Lever... Grinder: KitchenAid Pro Line Burr... Vac Pot: Vintage Silex Drip: Aeropress, French Press Roaster: "EL SUPREMO" w/QuikSPIN-CRC...
Posted Tue Jun 24, 2008, 9:30pm Subject: Re: Formula needed
Jordan,
A motor I now use is rated at 30 inch pounds of torque, 62 RPM. In relation to the result of the equation, it would appear to work, or would you think that the rating provided for the motor may have a different meaning?
senorswiss Senior Member Joined: 10 Feb 2008 Posts: 109 Location: West Lafayette, Indiana, usa Expertise: I live coffee
Espresso: Silvia Grinder: espresso: EB, others:Maestro... Vac Pot: Moka Pot Drip: French Press, Aero Press Roaster: Modified HG/BM
Posted Wed Jun 25, 2008, 5:24pm Subject: Re: Formula needed
Len,
I suppose it would be powerful enough to turn the cylinder, but it's not clear to me if it will be too powerful. Are you going to control the speed somehow or just use the motor at full power?
Don't forget that the moment of inertia I gave (.5m*r^2) is for a solid cylinder. If it's hollow that means all 60kg is around the edge of the cylinder, and it will thus require more torque to turn it.
Again, there are tables with moments of inertia for various shapes all over the internet.
Do you know what the torque is from 0 rpm to whatever speed it is going to eventually turn? If the torque at the low end is less than the torque require to spin the cyl. it's never going to start spinning.
I don't have any real working knowledge of designing these systems, so it's possible I am not catching everything. All I know is how to solve for torque. :)
CoffeeRoastersClub Senior Member Joined: 6 Jul 2005 Posts: 2,144 Location: Vernon Expertise: Professional
Espresso: Vintage La Pavoni Lever... Grinder: KitchenAid Pro Line Burr... Vac Pot: Vintage Silex Drip: Aeropress, French Press Roaster: "EL SUPREMO" w/QuikSPIN-CRC...
Posted Wed Jun 25, 2008, 5:53pm Subject: Re: Formula needed
Hello Jordan,
It is a cylinder, but has a good portion of the 60 pound weight I reference as beans (which start out as dead weight, then start to "float" quite a bit at the 60 rpm speed). Adds another factor to the equation likely.
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