Question about forces in a gearbox

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    Electric motor Forces
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SUMMARY

The discussion centers on calculating the output torque of a worm gearbox with a 10:1 gear ratio when powered by a 1 kW electric motor operating at 1500 RPM. The initial confusion arises from the misconception that the gearbox can increase power output, which is incorrect; power remains constant at 1 kW. The correct output torque is derived from the input torque, which is calculated as 6.37 N.m, multiplied by the gear ratio, resulting in an output torque of 63.7 N.m at 150 RPM. The efficiency of the gearbox, typically between 80-90%, must also be considered, as real-world applications will yield slightly less than the theoretical output.

PREREQUISITES
  • Understanding of basic mechanical principles, including torque and power.
  • Familiarity with gear ratios and their impact on speed and torque.
  • Knowledge of efficiency calculations in mechanical systems.
  • Ability to use power-torque calculators for verification of calculations.
NEXT STEPS
  • Learn about the efficiency of worm gearboxes and how it affects output power.
  • Study the relationship between torque, speed, and power using the equation P = τω.
  • Explore practical applications of gear ratios in various mechanical systems.
  • Investigate the differences between theoretical and actual performance in mechanical systems.
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Mechanical engineers, robotics enthusiasts, and anyone involved in designing or analyzing systems that utilize electric motors and gearboxes will benefit from this discussion.

Kiko
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hi all
i have a simple question but answer i don't know how to find :)

lets say we have a electric motor with 1kw at 1500 rpm and that is 4.75 N.m
when the electric motor is connected to worm gearbox n proportion 10:1 then the speed on output point of the gearbox will be 150 RPM. But this mean also that the torque on output point of the gearbox is increased, and that is 9.x kw, let's say 9 kw.
In this case, when we have 9 kw on output point of the gearbox with output speed 150 rpm then: output torque will be 427 N.m

i am correct or?!
If not, then let me know how much torque on N.m i will have on output point of the worm gearbox 10:1 when source el.motor is working on 1 kw with speed 1500 rpm.

im based on calculations on http://wentec.com/unipower/calculators/power_torque.asp

thank you
 
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Kiko said:
hi all
i have a simple question but answer i don't know how to find :)

lets say we have a electric motor with 1kw at 1500 rpm and that is 4.75 N.m
I used the calculator at the site you linked, and I got 6.37 N.m when I input 1kw and 1500 rpm.

when the electric motor is connected to worm gearbox n proportion 10:1 then the speed on output point of the gearbox will be 150 RPM. But this mean also that the torque on output point of the gearbox is increased, and that is 9.x kw, let's say 9 kw.
Here, I think, is the root of your confusion. Changing the gear ratio changes torque, not power. If you have a 1kw motor, you will get 1 kw out the end, regardless of gearing. So, try figuring your output torque again, bearing in mind that the horse power (or kw, in this case) will not be changed. I think you’ll like it, because the math is actually easier than you have been doing.
 
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LURCH said:
Here, I think, is the root of your confusion. Changing the gear ratio changes torque, not power. If you have a 1kw motor, you will get 1 kw out the end, regardless of gearing. So, try figuring your output torque again, bearing in mind that the horse power (or kw, in this case) will not be changed. I think you’ll like it, because the math is actually easier than you have been doing.

ok
then let me know how much torque on N.m i will have on output point of the worm gearbox 10:1 when source el.motor is working on 1 kw with speed 1500 rpm.
 
Kiko said:
ok
then let me know how much torque on N.m i will have on output point of the worm gearbox 10:1 when source el.motor is working on 1 kw with speed 1500 rpm.
Use the calculator to get the input torque. Multiply it by your gear ratio.
 
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@Kiko You need to be applying the equations that you already know in the right way or the answers will not be valid. Your initial calculation implies more power out than power in and you have to smell a rat in that answer.
Input torque will be (Input Power) / (input angular speed)
Ideally, Output torque and output speed will be scaled according to the gear ratio. Output Power will be (output torque) x (output angular speed)
That is a very theoretical answer, of course. A worm gear is very inefficient (which is why they don't work backwards) and the output Power will be significantly less than the input Power.
Basic theory about Machines in general uses two quantities.
Velocity Ratio - which is the ratio of the input and output speeds and which is just based on the geometry of the machine.
Mechanical Advantage - which is the actual ratio of forces or torques at output and input.
Efficiency = MA/VR (always less than one, of course)
People tend to use just Velocity Ratio in their calculations and then they get over optimistic ideas about the likely performance of their machine. Worse still - they refer to that as Mechanical Advantage. You will even find that practice all over PF threads. Tut tut.
 
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@sophiecentaur

ok, thanks for your answer, but nothing i understand here... sorry about that. But my question is simple and concrete, also i need simple and concrete answer, sure, if you would like :)
how much torque on N.m i will have on output point of the worm gearbox 10:1 when source el.motor is working on 1 kw with speed 1500 rpm?
thanks
 
Kiko said:
@sophiecentaur

ok, thanks for your answer, but nothing i understand here... sorry about that. But my question is simple and concrete, also i need simple and concrete answer, sure, if you would like :)
how much torque on N.m i will have on output point of the worm gearbox 10:1 when source el.motor is working on 1 kw with speed 1500 rpm?
thanks
Multiply your input torque by the gear ratio.
 
Kiko said:
hi all
i have a simple question but answer i don't know how to find :)

lets say we have a electric motor with 1kw at 1500 rpm and that is 4.75 N.m
when the electric motor is connected to worm gearbox n proportion 10:1 then the speed on output point of the gearbox will be 150 RPM. But this mean also that the torque on output point of the gearbox is increased..

Correct. The rpm reduces by 1/10. The torque is increased by *10 so 47.5Nm.

and that is 9.x kw, let's say 9 kw.

No you made a mistake somewhere. No gearbox can produce more energy out than it consumes.

If there are no friction losses in the gearbox the power output will be unchanged (eg 1kw in = 1kw out). However real gearboxes aren't 100% efficient. I think typical worm drive gearboxes are 80-90% efficient so you should only expect 800-900W output with 1000W input.
 
Kiko said:
hi all
i have a simple question but answer i don't know how to find :)

lets say we have a electric motor with 1kw at 1500 rpm and that is 4.75 N.m.

Actually 4.75Nm at 1500rpm isn't 1000W...

1500rpm = 157rads/s
157*4.75=740W

So did you mean a 1HP motor? Eg A motor that delivers 750W? That might consume 1000W if its only 75% efficient.
 
  • #10
CWatters said:
Actually 4.75Nm at 1500rpm isn't 1000W...

1500rpm = 157rads/s
157*4.75=740W

So did you mean a 1HP motor? Eg A motor that delivers 750W? That might consume 1000W if its only 75% efficient.

look like you are correct. Great! thanks
1 kw = 1.34 HP

my main goal is to mount the 1 kw electromotor to the worm gearbox 10:1 like source of power.
1 kw = 6.37 Nm torque
then how much torque will be n output point of the worm gearbox 10:1?

thanks CWatters
 
  • #11
Kiko said:
1 kw = 6.37 Nm torque
Only at one operating speed.
You have already been over (more than once) how torque and speed are related, for a given power. So, if the speed is divided by ten in the worm gear, you can say what happens to the torque. You do not actually need to be told that explicitly, do you?
 
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  • #12
Kiko said:
1 kw = 6.37 Nm torque
then how much torque will be n output point of the worm gearbox 10:1?
Please tell me you can multiply or divide a number by 10 in your head!
 
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  • #15
Kiko said:
look like you are correct. Great! thanks
1 kw = 1.34 HP

my main goal is to mount the 1 kw electromotor to the worm gearbox 10:1 like source of power.
1 kw = 6.37 Nm torque
then how much torque will be n output point of the worm gearbox 10:1?
I don’t think anyone has mentioned this to you yet, but people in these Forums will almost never just spit out an answer like that. It wouldn’t exactly be violating any rules, but it would be frowned upon; sort-of goes against the overall mission of this site, which is to increase scientific understanding. We are all trying to show you the right direction to find the answer yourself, and point out any missteps we can see that are keeping you from finding it.

Just thought you should know, nobody’s trying to be evasive or rude, just trying to stay true to our mission. So, take your best shot at what you think the answer is, and why, and we will help as much as we can. Based on what you’ve said so far, it looks like you probably know the answer.
 
Last edited:
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  • #16
It's general rule that, if one wants specialist information that can be relied on, it will cost you money because you will need to pay a professional for it. So many people who ask for information on PF seem to think they are 'customers' with customer rights. Strange when you think of the hours of input that a well presented appeal for help from a can elicit from PF when the questioner is prepared to make a bit of an effort.
 
  • #17
Ok, that's enough. Thread locked.

Edit by Mark44 -- A number of substandard posts and rants have been deleted.
 
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