Hi Gene, Quite right. That link I posted used a table while in fact, as you pointed out, the mass is mostly on the outside of a flywheel with spokes. I would imagine at there is some average where if it's a 300 lb disk that is 24" might be the same as a 36" disk that is 400 lbs.
Think of a fly press for example with a clutch that engages the tooling. Even if it does take 5 seconds to get up to speed, the clutch engages, the tool moves down and punches and moves up and the clutch releases. Even if the speed slowed down by 20% when the clutch released then assuming linear acceleration now only 1 second is required to bring the speed back up. At 50 RPM (0.83 seconds per rev) then you could do another punch stroke 1 second later and so possibly run 30 strokes per minute. That jpg chart I included suggests with 100% efficiency and no real friction that 45 oz-in are required. Seems very low to me hence the questions. Even if I did use a stepper motor and went 16:1 to bring the RPM down to 800 RPM the motor could easily be a size 23 300 oz-in. Could that actually bring a flywheel up to that speed in 5 seconds? > -----Original Message----- > From: gene heskett [mailto:ghesk...@shentel.net] > Sent: June-16-22 6:34 PM > To: emc-users@lists.sourceforge.net > Subject: Re: [Emc-users] Acceleration question. > > On 6/16/22 20:54, John Dammeyer wrote: > > OK. I realize this will be a dumb question but please bear with me > > especially since I've included the ability > to accelerate in my Electronic Lead Screw project. > > > > A friend and I were discussing bringing a 300 pound flywheel up to speed. > > Vz=0 RPM, Vf=50 RPM. Reduction drive to the flywheel shaft is 32:1 so > > final speed of motor is 1600 RPM. > > > > Assume we're happy with 5 seconds to accelerate for Tz to Tf. Motor > > voltage is 12V. > > > > We have the mass, we have the velocity, we have the time and motor voltage. > > The question is what are > the calculations to determine how much current the motor will require to > create this acceleration? > Assuming of course the motor is 100% efficient. > > > > We're getting all confused with F=ma and 1/2*a*t^2 etc. > > > > What size motor is actually needed to do this? > > > > Thanks. > > John > > > That John, is going to be determined by where that weight is. > If 270 lbs of it is in a rim 4 feet in diameter and the other 30 > is in the spokes supporting that rim, its going to take a lot > more torque to get it up to speed in 5 seconds than it would > take if its only 2 feet in diameter, its the linear speed of the > outer diameters major mass that has to be moved to twice > as many feet per second needing 4x the torque to do it for > the 4 foot example, and Einstiens E=m*v*v comes into the > picture, cuz v=2*2 is 4, but v=4*4 is 16, not 8. > > That's as close as I can get to the math, sorry. I'd have to > ask someone else for a SWAG or more knowledgeable > answer too. This is a case also, of doing a bit of cheating > with a bigger vfd running at a higher voltage and the low > speed current boost could, if enough line voltage is present, > bang a 1 horse motor hard enough to natch a 3 or 4 horse > motor, knowing the overdrive will only last a few seconds. > > But, if going to machine cut with that motor, I'd have an > amprobe or equ watching the motor currant to make sure > the steady load is within the FLA on the motors nameplate. > > I hope the real answer means you've a motor and vfd in > stock that will do it. > > Cheers, Gene Heskett. > -- > "There are four boxes to be used in defense of liberty: > soap, ballot, jury, and ammo. Please use in that order." > -Ed Howdershelt (Author, 1940) > If we desire respect for the law, we must first make the law respectable. > - Louis D. Brandeis > > > > _______________________________________________ > Emc-users mailing list > Emc-users@lists.sourceforge.net > https://lists.sourceforge.net/lists/listinfo/emc-users _______________________________________________ Emc-users mailing list Emc-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/emc-users
