Originally Posted by
The Blight
I'm just checking if my calculations are correct (I'm just taking the logical aproach here).
I want to know how much force you need to hold a tool in a MT3 collet, and so I did some calculations using some numbers on how much force was needed on a R8 collet.
The R8 collet has a ratio of 1:7
Some dimensions of the R8
D1=31,75
D2=24,13
L=26,67
31,75-24,13=7,62
7,62/2=3,81
3,81/26,67=0,142857
0,142857=1/7
The MT3 collet has a ratio of 7/279
Some dimension of the MT3
D1=23,825
D2=19,7612
L=80,9752
23,825-19,7612=4,0638
4,0638/2=2,0319
2,0319/80,9752=0,02509
0,02509=7/279
The R8 need about 1500lbs of force to clamp the tool securly in the spindle. This gives:
1500=680Kg
680/0,142857(Ratio of R8)=4760Kg
To get this kind of force on a MT3:
4760*0,02509(ratio of MT3)=119Kg
I know there are some factors that are not taken into this calculation, but it gives me some clue as to how much force I need to apply to the MT3 drawbar.
I have seen the number 1000lbs for the R8 too in a topic, and this gives about 80Kg on the MT3.
With only 80Kg of force needed, I could use a 200oz-in stepper connected to a 1mm pitch screw to push the drawbar down (pushed upwards by spring washers) as it would give 200*28=5600, 5600*25,4=142240g / 142Kg. If I made the stepper with a 1:4 ratio, I could use a 60oz/in stepper (want to keep the size of this thing down).
So are these calculations right?