carlosiltopochegira
Guest
OK thanks to everyone, now I try to do the exercise and then I share everything
good morning to all, can someone help me in this design? i find myself with a professor who cannot help me do this exercise
carlosiltopochegira
Guest
meccanicamg
Guest
I will be tired after a whole day at work but:
What's the acceleration to?
- 270° : 5s = 360° : x ....make the reverse formula and I find that x=6.67 seconds/turn
- to have the spins/minute that is n you do simply 60minuti/second divided 6.67 seconds/turn to find 9 turns/minute
- to obtain omega ....2*pigreco*n / 60= 0.942 rad/s
Why do you make up the formulas? are the usual formulas of number laps, angular speed that are always used and everywhere, at work and at school.
equal torque multiplied angular speed....40nm*0.942=37.7w which divides the 0.9 yield get an input power of 41.87w.
Why is the degree of breaking safety worth 3.25? Why so and not 3 or 6?
It's all in the wrong waterfall.
c45 normalized or bonified expanse, which has the highest breaking load of the c15 charcoal or wired or hardened to heart? then always indicates that treatment you do it and the basic material if it is filled, tempered at heart etc. do not be summary.
Also in the last form consider 40nm....but if you lose power and torque....you must insert 40/0.9=44.5nm
then write: "I'll have m=2..." and then you'll come to calculate "m=0.6". conclud....write 0.6 less than 2, positive verification. do not throw in numbers and formulas.
and you even had different tricks before, dispenses, calculations, etc...
What's the acceleration to?
- 270° : 5s = 360° : x ....make the reverse formula and I find that x=6.67 seconds/turn
- to have the spins/minute that is n you do simply 60minuti/second divided 6.67 seconds/turn to find 9 turns/minute
- to obtain omega ....2*pigreco*n / 60= 0.942 rad/s
Why do you make up the formulas? are the usual formulas of number laps, angular speed that are always used and everywhere, at work and at school.
equal torque multiplied angular speed....40nm*0.942=37.7w which divides the 0.9 yield get an input power of 41.87w.
Why is the degree of breaking safety worth 3.25? Why so and not 3 or 6?
It's all in the wrong waterfall.
c45 normalized or bonified expanse, which has the highest breaking load of the c15 charcoal or wired or hardened to heart? then always indicates that treatment you do it and the basic material if it is filled, tempered at heart etc. do not be summary.
Also in the last form consider 40nm....but if you lose power and torque....you must insert 40/0.9=44.5nm
then write: "I'll have m=2..." and then you'll come to calculate "m=0.6". conclud....write 0.6 less than 2, positive verification. do not throw in numbers and formulas.
and you even had different tricks before, dispenses, calculations, etc...
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meccanicamg
Guest
I downloaded to try the khk catalog of a rack but reports the strength and not the pair.
PIETRO2002
Guest
try to select in the following way:
selections : select gear
then select : gear rack
At this point you choose a steel rack, it will appear in the central part of the screen the design of the rack and on the right side the bearable torque values in nm, probably it's a mistake, I didn't make up anything. .
meccanicamg
Guest
I checked. in the rack selection configurator there is the standard wording of the gears that is expressed in nm.... but it was in n. the paper catalog in fact reports right, same number but n......forza is ???