25 Hz traction power system (Rail transit)
- Thread starter tortuga
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pg 67 on https://www.rexnord.com/contentitems/techlibrary/documents/131-110_catalog
input power ~1000 hp
1750 rpm
12.61:1 size 465 1165 hp wt 5400 lbs
720 rpm
5.06:1 size 445 1023 hp wt 4320 lbs
double reduction
as stages of reduction increase so does the wt difference
input power ~1000 hp
1750 rpm
12.61:1 size 465 1165 hp wt 5400 lbs
720 rpm
5.06:1 size 445 1023 hp wt 4320 lbs
double reduction
as stages of reduction increase so does the wt difference
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yes, I estimated those values
but you do not understand their relevance lol
the output torque or input power determine the size
the torque passed thru the box is determined by the load
same vehicle wt and speed = same load
Continue to denigrate me. Fine if that's what floats your boat. Just stick to facts
you do the same
'I would expect you to understand' etc
backhanded, mine are more direct
those are facts
when adjusted for ratio
the 445 4320 lbs
465 5750 lbs
that holds true for all gearboxes
same input hp
same output torque
the lower the numerical ratio the lower the wt
this may help
1000 hp
1800 rpm, 3000 lb ft, 4:1, 12000 lb ft, 450 rpm
900 rpm, 6000 lb ft, 2:1, 12000 lb ft, 450 rpm
the output/ring gears are in synch/same speed
assume r = 1' for math ease
the size doesn't matter as long as ratios are maintained
radius r is ~ tooth count/ratio
1800 input/pinion r = 1/4
900 r = 1/2
T = F x r or F = T/r
1800: F = 3000/(1/4) = 12000 lb (force)
900: F = 6000/(1/2) = 12000
the force trying to twist the shaft and on the gear teeth is the same
that is why for a given input power, output torque and speed, the shaft will see the same stress regardless of INPUT torque
the shaft trasmits torque and is subject to the force
1000 hp
1800 rpm, 3000 lb ft, 4:1, 12000 lb ft, 450 rpm
900 rpm, 6000 lb ft, 2:1, 12000 lb ft, 450 rpm
the output/ring gears are in synch/same speed
assume r = 1' for math ease
the size doesn't matter as long as ratios are maintained
radius r is ~ tooth count/ratio
1800 input/pinion r = 1/4
900 r = 1/2
T = F x r or F = T/r
1800: F = 3000/(1/4) = 12000 lb (force)
900: F = 6000/(1/2) = 12000
the force trying to twist the shaft and on the gear teeth is the same
that is why for a given input power, output torque and speed, the shaft will see the same stress regardless of INPUT torque
the shaft trasmits torque and is subject to the force
I prefer P=Tω.
Phil Corso
Senior Member
- Location
- Boca Raton, Fl, USA
Gentlepeople?
Just to change obvious misdirection of this topic, my 1st job with a Sheetrock mfg used 25Hz Solenoids to vibrate storage bins and hoppers!
Regards, Phil Corso
Just to change obvious misdirection of this topic, my 1st job with a Sheetrock mfg used 25Hz Solenoids to vibrate storage bins and hoppers!
Regards, Phil Corso
that is because you don't understand
your assertion has been proven wrong
the lower ratio is lighter
I prefer T = F X r
P = Tn does not show why the stress is equal for both ratios when power is equal
one glance in the catalog shows that for a given power and speed output the shaft diameter is the same regardless of torque input, ie, ratio
so, the lower ratio/higher T input is heavier?
pg 67 on https://www.rexnord.com/contentitems...31-110_catalog
input power ~1000 hp
1750 rpm
12.61:1 size 465 1165 hp wt 5400 lbs
720 rpm
5.06:1 size 445 1023 hp wt 4320 lbs
nope
low speed size 445
high speed size 465, heavier, larger shaft
same input power, same output speed/torque
winnie
Senior Member
- Location
- Springfield, MA, USA
- Occupation
- Electric motor research
On the discussion of gearbox mass, I believe that the following hold:
The largest torque dominates the mass; in the case of a speed reducer this is the _output_ torque.
The smaller the gear ratio, in general, the lower the total mass of the gearbox.
The higher the input torque the higher the mass of the input components.
I suspect that for the same number of reduction stages, a larger gear ratio (meaning lower input torque) will mean lower total mass, bucking the general trend.
-Jon
The largest torque dominates the mass; in the case of a speed reducer this is the _output_ torque.
The smaller the gear ratio, in general, the lower the total mass of the gearbox.
The higher the input torque the higher the mass of the input components.
I suspect that for the same number of reduction stages, a larger gear ratio (meaning lower input torque) will mean lower total mass, bucking the general trend.
-Jon
So I should not possibly have designed paper mill rewinder drives without understanding the varying inertia, tensions? , coiling machinesin steel mills?
Rolling mill drives? Because I don't understand?
Somehow I got by. For 50 years.
And continually got repeat business.
It's because I do.
at least you apparently think you do
look at the catalog
for constant power, output torque and output speed, the lower ratio will be lighter
not heavier as you asserted while attempting to correct me, a hobby of yours
simple solution: avoid each other
OK. You win. You found one example that supports your point. Except the motor, the input shaft, the coupling, the bed plate and mounting plate etc will all be heavier. But never mind that.
And I admit that my experience with kit for rail transit (the OPs topic) is limited to those countries countries connected by the VSOE and their various voltages and frequencies.
Oh well, at least I have enjoyed a glass of Champagne on the Orient Express - and, to paraphrase Michael Caine, not a lot of us have done that.....................
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rlundsrud
Senior Member
- Location
- chicago, il, USA
I swear, I am going to turn this car around and then no chucky cheese for anyone!!
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