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 Help regarding gear calculations |
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| Help regarding gear calculations |
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Robse
Mitglied
Dabei seit: 24.09.2011
Beiträge: 31
Maßstab: 1:14,5
Wohnort: Dänemark
Level: 27 [?]
Erfahrungspunkte: 142.338
Nächster Level: 157.092
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Hello,
Please excuse me for writing in english, but my german is not good enough yet.
My question concerns the built of my LTM1500-8.1 in 1:14 scale. I have reached the point where I need to build / find a gear box and motor.
The track where I will be running this crane has roads with climbs up to 10 degrees. I therefore want the crane to be able to run on 10* slobes.
As a test I put the chassis with axles etc, and weights so that the total weight was 41 kg's on a 10 degree ramp, and then mesured the force needed to pull the whole thing up hill: 10 kg of pulling force.
My wheels are 10,8cm in diameter, and my 8 TLT axles all have individual differentials with a gearing of 2,67:1
I have tried to calculate how much force (Ncm) that I must apply to the cardan shaft to equal 10kg's of pulling force.
My result is 20,2 kg/cm, or 198Ncm... but I am unable to double check this.
How did I calculate it?
I first calculated the force needed as if the crane had only one wheel. I calculated the F=m*a for the wheel: F = 10kg * 5,4cm
That's equal to 54kgcm on all of the the wheel shafts in total.
I then divided that by the gearing in the differential: 54kgcm / 2,76 = 20,2 kgcm. That tells me that the needed torque after the gearbox is 20kgcm, or close to 200Ncm, on the cardan shaft.
(The number of wheels sharing that force does not matter, I think. As long as all of the force ends up on the road, all is ok. I therefore look at it as if there were only one wheel.)
Or does it... can anyone confirm my method, please? How do you make sure that you have enough power from engine, gearbox(es)?
Thank you very much in advance, guys!
Robert Holsting, Denmark
robse.dk
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19.05.2013 19:56 |
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Robse
Mitglied
Dabei seit: 24.09.2011
Beiträge: 31
Maßstab: 1:14,5
Wohnort: Dänemark
Level: 27 [?]
Erfahrungspunkte: 142.338
Nächster Level: 157.092
Themenstarter 
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Additional info: Why 41kg of test weight? Because that is what the crane will end up at, according to my estimate.
A small image of the test can be seen here:
The whole project can be seen at robse.dk :-)
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19.05.2013 20:09 |
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| RE: Help regarding gear calculations |
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Kette1zu8
Foren Gott
Dabei seit: 24.08.2011
Beiträge: 2.218
Maßstab: 1:10
Wohnort: Bayern
Level: 50 [?]
Erfahrungspunkte: 10.253.742
Nächster Level: 11.777.899
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Hi there,
It takes 10Kg to pull the crane up the 10° slope, so that is a force of 100N. Wheel radius is 0.054m; total torque is 5.4Nm, so your first calculation was right.
If the whole load goes through only one axle:
5.4Nm : 2.67ratio = 2.02Nm --> you're right again!
For the right motor size you should tell us the maximum speed of the model.
__________________
Gruß, Markus
Liebherr R984 Litronic im Bau, M1:10
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19.05.2013 20:43 |
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Robse
Mitglied
Dabei seit: 24.09.2011
Beiträge: 31
Maßstab: 1:14,5
Wohnort: Dänemark
Level: 27 [?]
Erfahrungspunkte: 142.338
Nächster Level: 157.092
Themenstarter
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Thank you very much! :-)
Am I also right that the number of axles that are sharing the force on the cardan shaft does not matter? I have eight axles, but they are all linked on the same cardan, sharing gearbox and engine.
Speed: I aim at 1km/h, but are willing to trade speed for torque.
I think I will take a Veroma gearbox, and use 1 and 2 gear. That's 11,3:1 and 22,6:1
Then, after the veroma, I will put a 5:1 reducing-box, that also have a shaft going in both directions. (The gearbox will "hook on" to the cardan axle between axle 6 and 7)
The motor will most likely be a Truckpuller 3, 12V model. (6300 rpm, 36Ncm)
Plenty of power! (I think....) :-)
Dieser Beitrag wurde 1 mal editiert, zum letzten Mal von Robse: 19.05.2013 21:05.
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19.05.2013 20:53 |
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Kette1zu8
Foren Gott
Dabei seit: 24.08.2011
Beiträge: 2.218
Maßstab: 1:10
Wohnort: Bayern
Level: 50 [?]
Erfahrungspunkte: 10.253.742
Nächster Level: 11.777.899
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yes, that's right. The whole load comes out from the gearbox and goes into the main shaft. It doesn't matter how many axles or wheels.
OK, let's calculate with 1Km/h (0.28m/s):
Wheel circumference = 0.108m x 3.14 = 0,34m
Wheel refs for 0.28m/s: 0.28m/ / 0,34m = 0.82 1/s = 49.2 1/min
Diff input shaft refs : 49.2 1/min x 2.67 = 131.4 1/min
131.4 1/min x 5 x 22.6 = approx. 14850 1/min motor refs (1 Km/h 1. gear)
131.4 1/min x 5 x 11.3 = approx. 7425 1/min motor refs (1 Km/h 2. gear) -->
Second gear will speed up to 0.85 Km/h; maximum motor torque: 3.57Ncm + efficiency. Should be no problem for a Truckpuller 3!
__________________
Gruß, Markus
Liebherr R984 Litronic im Bau, M1:10
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19.05.2013 21:25 |
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Robse
Mitglied
Dabei seit: 24.09.2011
Beiträge: 31
Maßstab: 1:14,5
Wohnort: Dänemark
Level: 27 [?]
Erfahrungspunkte: 142.338
Nächster Level: 157.092
Themenstarter
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Excellent.
Thank you very much for taking the time to check, and answer. :-)
This is my first serious encounter with gears, and I would hate to get it wrong.
Best regards,
Robert Holsting, Denmark
robse.dk
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19.05.2013 21:38 |
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Kette1zu8
Foren Gott
Dabei seit: 24.08.2011
Beiträge: 2.218
Maßstab: 1:10
Wohnort: Bayern
Level: 50 [?]
Erfahrungspunkte: 10.253.742
Nächster Level: 11.777.899
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You're welcome!
__________________
Gruß, Markus
Liebherr R984 Litronic im Bau, M1:10
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19.05.2013 21:41 |
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JensR
Administrator
Dabei seit: 16.01.2010
Beiträge: 5.265
Maßstab: 1:14,5
Wohnort: England
Level: 56 [?]
Erfahrungspunkte: 27.422.072
Nächster Level: 30.430.899
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Just as a different way of approaching this:
41 kg on a 10° slope give a longitudinal force of 41kg*9.81m/s^2*sin(10°)=70N
To calculate the necessary power, multiply by the speed in km/h and divide by 3.6
For Markus' assumption of 1 km/h, that gives
70 N * 1/3.6 m/s = 20W
Not including any losses, or rolling resistance.
If I understood you correctly, you measured the force needed to pull it up, not the load to keep it at a height.
In that case, the difference between your 100N and my 70N would be the rolling resistance.
100N gives around 28W power for the 1km/h. (Again without gearbox losses)
What I dislike about LRP is that they do not really tell a lot about the motor. Has anyone found a decent datasheet?
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20.05.2013 00:17 |
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Kette1zu8
Foren Gott
Dabei seit: 24.08.2011
Beiträge: 2.218
Maßstab: 1:10
Wohnort: Bayern
Level: 50 [?]
Erfahrungspunkte: 10.253.742
Nächster Level: 11.777.899
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Just to confirm Jens' answer:
due to the truckpuller's rpm there will be a speed of 0.85 Km/h. Using 100N for rimpull to go up the 10° slope it requires 23Watt. I calculated this using the product of motortorque and motorrefs divided by 9550. Equal is in KW. Also, not included efficiency.
I didn't know this easy calculating, so again what learned!
@Jens
zum Truckpuller3 kann ich leider nichts beitragen, nur das bekannte Datasheet zum TP2;
__________________
Gruß, Markus
Liebherr R984 Litronic im Bau, M1:10
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20.05.2013 01:01 |
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JensR
Administrator
Dabei seit: 16.01.2010
Beiträge: 5.265
Maßstab: 1:14,5
Wohnort: England
Level: 56 [?]
Erfahrungspunkte: 27.422.072
Nächster Level: 30.430.899
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Power = force*speed = torque * rotational speed = pressure * flow rate = current * voltage
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20.05.2013 01:34 |
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Robse
Mitglied
Dabei seit: 24.09.2011
Beiträge: 31
Maßstab: 1:14,5
Wohnort: Dänemark
Level: 27 [?]
Erfahrungspunkte: 142.338
Nächster Level: 157.092
Themenstarter
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Hello,
Yes, the 10kg of force was needed to start and maintain forward motion, on the 10* slope. I did (for the fun of it :-) ) measure the force needed to hold a position, and got that to around 5kg's.
My gauge is a mechanical one, so it is not that accurate.
The data on the Truckpuller 3 has been picked from my favorite shop in Germany: Der Getriebedoctor. I have also confirmed the data by Google'ing it to other websites, and comparing data. Rpm @ 12V should be 6300, and torque should be app. 36Ncm. The high torque is what cought my eye in my search, and the 540 casing.
I hope that one day it will be standard to put all engine data, and not only the voltage, length, width, and weight. More important is rpm, torque and power needed (A) but this is often hard to find.
Thank you both for the additional input :-)
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20.05.2013 16:33 |
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Robse
Mitglied
Dabei seit: 24.09.2011
Beiträge: 31
Maßstab: 1:14,5
Wohnort: Dänemark
Level: 27 [?]
Erfahrungspunkte: 142.338
Nächster Level: 157.092
Themenstarter
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Oh, a little more on engine choice:
I was looking at some brushless outrunners because the effenciency is much, much better than ordinary brushed engine, but I went against it at the end.
Why? Well... My crane will drive around only very little. Most of the time it will be setup at a site, working (or at least simulating...hehe)
Second, I really like the way that the Servonaut M20+ handles motor speed, which is quite different from "normal" motor controllers.
For those not familar with Servonaut M20+, please let me explain:
When you move the stick forward, you add speed. When you pull the stick back, you remove speed (brake). In the center position the motor maintain speed that it currently has!
That enables me to set the speed, and then concentrate on steering, lights, blinkers, horn and what not.. and not worry about going too fast or soo slow all of a sudden. :-)
Also a great plus is that the basic light functions are build in, such as reverse, blinkers, brake light, position and low beam.
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20.05.2013 16:59 |
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Robse
Mitglied
Dabei seit: 24.09.2011
Beiträge: 31
Maßstab: 1:14,5
Wohnort: Dänemark
Level: 27 [?]
Erfahrungspunkte: 142.338
Nächster Level: 157.092
Themenstarter
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22.05.2013 20:15 |
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JensR
Administrator
Dabei seit: 16.01.2010
Beiträge: 5.265
Maßstab: 1:14,5
Wohnort: England
Level: 56 [?]
Erfahrungspunkte: 27.422.072
Nächster Level: 30.430.899
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One thing: in English "engine" refers to an internal combustion engine or a steam engine (=external combustion engine). Electric and hydraulic machines are called motors.
The tool looks good, I haven't checked the calculations, though.
I think you could make the description more clear:
Input minimum pulling force of the vehicle
Input minimum max speed
Unless I misunderstand what you are asking, maybe this would be clearer:
Input pulling force at minimum acceptable speed (up a ramp, for example)
Input minimum acceptable speed
Or you might think about even more detail
mass
rolling resistance
desired ramp angle
desired speed up the ramp
desired top speed on flat
...
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23.05.2013 00:42 |
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Robse
Mitglied
Dabei seit: 24.09.2011
Beiträge: 31
Maßstab: 1:14,5
Wohnort: Dänemark
Level: 27 [?]
Erfahrungspunkte: 142.338
Nächster Level: 157.092
Themenstarter
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Hi Jens,
Hmm, good one!
I will look into that, and improve the tool. Thanks :-)
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23.05.2013 10:07 |
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Robse
Mitglied
Dabei seit: 24.09.2011
Beiträge: 31
Maßstab: 1:14,5
Wohnort: Dänemark
Level: 27 [?]
Erfahrungspunkte: 142.338
Nächster Level: 157.092
Themenstarter
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26.05.2013 16:02 |
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