Question about forces in a gearbox

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    Electric motor Forces
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Discussion Overview

The discussion revolves around calculating the output torque of a worm gearbox connected to an electric motor, specifically focusing on the relationship between input power, torque, and gear ratios. Participants explore the implications of a 10:1 gear ratio on the output torque and power, while addressing potential misunderstandings regarding power conservation in mechanical systems.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant states that with a 1 kW motor at 1500 RPM, the input torque is approximately 4.75 N.m, and questions the output torque after connecting to a 10:1 worm gearbox.
  • Another participant clarifies that changing the gear ratio affects torque but not power, asserting that the output power remains 1 kW regardless of the gearbox configuration.
  • Some participants suggest recalculating the output torque by considering the input torque and the gear ratio, emphasizing that the output torque should be the input torque multiplied by the gear ratio.
  • There are discussions about the efficiency of worm gearboxes, with some noting that real-world efficiency is typically less than 100%, which affects the output power.
  • Participants also debate the initial calculation of torque and power, with one pointing out that the stated torque at the given RPM does not correspond to 1 kW, suggesting a possible misunderstanding of the motor's specifications.
  • One participant expresses frustration over the complexity of the explanations and requests a straightforward answer regarding the output torque.
  • Another participant emphasizes the importance of understanding the relationship between torque, speed, and power, indicating that these concepts have been discussed multiple times in the thread.

Areas of Agreement / Disagreement

Participants generally agree that the output power of the gearbox cannot exceed the input power, but there is disagreement on the specific calculations and interpretations of torque and power relationships. The discussion remains unresolved regarding the exact output torque value, with multiple viewpoints presented.

Contextual Notes

There are limitations in the assumptions made regarding the efficiency of the gearbox and the accuracy of the initial torque calculations. Some participants challenge the initial torque values and the implications of the gear ratio without reaching a consensus on the correct output torque.

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.
 
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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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