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Force generated by rotating mass 1
- Thread starter Ralph2
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- #21
There are quite a few styles out there. I have never seen the machine that holds the rotor in question but here is a link to something similar A search for brush mulcher will bring up quite a few samples.
The original question / discussion arose.. Because the customer wanted it balanced SANS teeth and... while in the process of balancing this rotor I wondered what the actual value (force) of the unbalance would be at the bearings.
The original question / discussion arose.. Because the customer wanted it balanced SANS teeth and... while in the process of balancing this rotor I wondered what the actual value (force) of the unbalance would be at the bearings.
You may add counterbalance weights to reduce the imbalance of the staggered teeth rows. Add the weights after the teeth are installed if you need to balance the rotor before adding the teeth. It would be a calculated balance method rather than a measure and balance method. But close enough to reduce the imbalance and feedback to the operator.
Ted
Ted
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- #23
Okay.. I think I am happy with my latest rational. I have two equal opposing forces acting on a rotor at (set 3/6) 32.625 and 41.75 inches from one end of my rotor which is 73.5 inches long.
Around one end I will have F(15321.54)@ D(73.5-32.625)/73.5=8515.648 Around the other end I will have 15312.54*41.75/73.5=8697.941
I do this for all 3 sets and add the results at each end giving me a net value of 675.59 and 631.48. A realistic value and is roughly equivalent to an ISO 1940 value of 60.
Have I gone wrong?? my calculations (spreadsheet) are attached
Around one end I will have F(15321.54)@ D(73.5-32.625)/73.5=8515.648 Around the other end I will have 15312.54*41.75/73.5=8697.941
I do this for all 3 sets and add the results at each end giving me a net value of 675.59 and 631.48. A realistic value and is roughly equivalent to an ISO 1940 value of 60.
Have I gone wrong?? my calculations (spreadsheet) are attached
I will make you less happy.
For your row set 3/6 example you did not take into account the moment caused by row 3 about the rh bearing, 15312*32.625/73.5 = 6797, in order to determine the lh bearing reaction 8698 - 6797 = 1901. Calculate 15312*(41.75-32.625)/73.5 for the bearing reaction of 1901 lbs. There are two moments around each bearing. One caused by each row of teeth. Each row generates a moment countered unequally by its opposing row resulting in a twisting couple which is the product of the force 15312 and the row center offset from each other. Divide this couple by the bearing spacing 73.5 to get the bearing reaction to that row's imbalance couple.
Combine each bearing reaction to arrive at the resultant reaction. Keep in mind the direction of all row set reactions when you calculate the resulting bearing reaction.
Ted
For your row set 3/6 example you did not take into account the moment caused by row 3 about the rh bearing, 15312*32.625/73.5 = 6797, in order to determine the lh bearing reaction 8698 - 6797 = 1901. Calculate 15312*(41.75-32.625)/73.5 for the bearing reaction of 1901 lbs. There are two moments around each bearing. One caused by each row of teeth. Each row generates a moment countered unequally by its opposing row resulting in a twisting couple which is the product of the force 15312 and the row center offset from each other. Divide this couple by the bearing spacing 73.5 to get the bearing reaction to that row's imbalance couple.
Combine each bearing reaction to arrive at the resultant reaction. Keep in mind the direction of all row set reactions when you calculate the resulting bearing reaction.
Ted
the bearing reaction will depend upon the stiffness of the supports. To develop any where near the calculated reaction force, the supports would have to be very stiff indeed. With the vehicle on its suspension the 2200 rpm may be well above some rigid body mode, and the vehicle (or rotor frame on flexible arms) motion will be mass controlled, not stiffness controlled. If so, the rotor will be rotating about its CG, and the bearing supports will be moving an amount related to the offset or eccentricity of the CG to the shaft mechanical center.
G60 would be on the order of 0.01 inch eccentricity, or 0.02 inch peak-to-peak or TIR.
See page 3 or 4 here -
G60 would be on the order of 0.01 inch eccentricity, or 0.02 inch peak-to-peak or TIR.
See page 3 or 4 here -
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- #26
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- #27
Okay.. re done and I hope for the last time. Am I right this time????
Set 1/4, the right side bearing would see a force of (37.125*15312.54=35.25*15312.54+73.5*X) =390.626 pounds. The left side bearing would see the same force (36.375*15312.54=38.25*15312.54+73.5*X) but as a negative value of -390.626
Set 2/5, RS (39*15312.54=33.5*15312.54+73.5*X) =1145.836 LS (34.5*15312.54=40*15312.54+73.5*X)
Set 3/6 RS, (40.875*15312.54=31.75*15312.54+73.5*X) = 1308.064 LS (32.625*15312.54=41.75*15312.54+73.5*X)
When these forces are combined the net result is a pull of 1308.064 pounds on one bearing with an equal pull in the opposite direction on the other bearing.
This works out roughly to an ISO value of G120.. I am pretty sure I will see this rotor back soon to balance with the teeth installed.
Set 1/4, the right side bearing would see a force of (37.125*15312.54=35.25*15312.54+73.5*X) =390.626 pounds. The left side bearing would see the same force (36.375*15312.54=38.25*15312.54+73.5*X) but as a negative value of -390.626
Set 2/5, RS (39*15312.54=33.5*15312.54+73.5*X) =1145.836 LS (34.5*15312.54=40*15312.54+73.5*X)
Set 3/6 RS, (40.875*15312.54=31.75*15312.54+73.5*X) = 1308.064 LS (32.625*15312.54=41.75*15312.54+73.5*X)
When these forces are combined the net result is a pull of 1308.064 pounds on one bearing with an equal pull in the opposite direction on the other bearing.
This works out roughly to an ISO value of G120.. I am pretty sure I will see this rotor back soon to balance with the teeth installed.
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- #29
Sorry I answered without looking at my work. The set calculations are the same except check your 3/6 set calculation. I think it should be 1900 lbs.
My resultant force at the right hand bearing is 2640 lbs. acting nearly 90 degrees clockwise from row 1. I assumed rows numbered 1 thru 6 in a counter-clockwise direction looking end on at the left hand bearing.
There are some strange keystrokes in your equations. Typos?
Ted
My resultant force at the right hand bearing is 2640 lbs. acting nearly 90 degrees clockwise from row 1. I assumed rows numbered 1 thru 6 in a counter-clockwise direction looking end on at the left hand bearing.
There are some strange keystrokes in your equations. Typos?
Ted
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- #31
Thanks Ted
You are quite right, I copied and pasted something that is different than the spreadsheet I was using. I do have 1901.047 as a result of set 3/6. From this computer I can not add the resulting forces so will take your word for the resulting 2640 pounds.
You are quite right, I copied and pasted something that is different than the spreadsheet I was using. I do have 1901.047 as a result of set 3/6. From this computer I can not add the resulting forces so will take your word for the resulting 2640 pounds.
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- #32
Ted, your resulting force (2640) is almost exactly double (1308.66) what a little DOS program (from IRD)I have used for years to combine weights. I used a drafting program (Solidworks) to calculate the results and agree with your figure (2638.710)...
I may have to retire my DOS program or at least be skeptical of the results.. Thanks for your help
I may have to retire my DOS program or at least be skeptical of the results.. Thanks for your help
All - I'm new to the forum, just ran across this topic on a google search today. I'm trying to recreate the original ~82,500lb force from the 80lb weight at r=7.5". The only way I can do so is to divide the 80lb weight by 9.8 m/s^2 - was there a unit error in the calculation? I believe the actual value is approx. 2090 lbf.
F = mrw^2
m = 80lb/32.17 ft/s^2 = 2.49 slug
r = 7.5 in = 0.625 feet
w = 2200 RPM/60 = 36.67 RPS
F = 2092 lbf
I may be way off base.
F = mrw^2
m = 80lb/32.17 ft/s^2 = 2.49 slug
r = 7.5 in = 0.625 feet
w = 2200 RPM/60 = 36.67 RPS
F = 2092 lbf
I may be way off base.
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- #36
The formula I use is somewhat bastardized to give the results in pounds of force while dealing with a weight in grams. So the 80 pounds ends up being 80*16*28.349=36286.72 gram. Then my formula .062*(N/1000)^2*radius*weight in grams .062*4.84*7.5*36286.72=81666.892 pounds of force.
How accurate the number is..is what led to the original question. I don't recall from where I got this (perhaps some balancing course). I believe if it gives an approximate value but only over a certain speed range... For example if one uses the speed of 10 RPM the value is only 3 pounds.. obviously not true
How accurate the number is..is what led to the original question. I don't recall from where I got this (perhaps some balancing course). I believe if it gives an approximate value but only over a certain speed range... For example if one uses the speed of 10 RPM the value is only 3 pounds.. obviously not true
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