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Torque to force conversion.

11/15/2006 11:23 PM

Hi! I need to move a shaft of 25 Kg over low friction ballbearings in a rectilinear fashion along its axis.

Considering a low co-eff of friction let us say that the force required is 35KgF or 350 N.

I have a motor delivering some amount of torque say Y Ncm.

Let the radius of the shaft be X

Now how do I compute the force that the torque would produce when applied on the head of the shaft.

Is there a formula that can be used here. I tried searching google but I never could understand how to interpret torque in terms of Force applied.

I understand that it is the basic engineering fundamentals but I am an electrical-electronics engineer and know a little mechanical but here I am stuck.

Please let me know.

If the question posted here is against the regulations of the forum please let me know that as well...

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#1

Re: Torque to force conversion.

11/16/2006 12:43 AM

Hi

Use this method

Torque = Force * Distance(Radius)

Nagaraja

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#10
In reply to #1

Re: Torque to force conversion.

11/18/2006 12:03 AM

Thank you Nagaraja,

I do understand that part of the equation but the problem is which radius or in reference to what do I need to consider for the calculation...

Thank you again for your feedback.

with regards,

Vishal Vijay Singh

(+919940131329)

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#2

Re: Torque to force conversion.

11/17/2006 2:57 AM

To use your variable name then;

Force = Torque ÷ Radius

Force = y/x

For example if your motor had a shaft on it with a 2cm diameter then to lift your 35Kg/350N weight;

Torque = Force x Radius = 350N x 0.01m = 3.5Nm

Conversely if your motor produced say 50Nm torque and you wished to lift the same weight then;

Radius = Torque ÷ Force = 50 ÷ 350 ≈ 0.143m = 143mm or 286mm Diameter

Keep in mind though that the usual units are Newton Meters Nm and multiples of it not Ncm as you stated in your question. 1 Ncm should be expressed as 0.01Nm or even better 10μNm. I know I am being pedantic but to restate my much used quote.

"It's important that we all use the same units as it minimizes the risks of misunderstandings like the one that caused the loss of a Mars explorer space craft."

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#3
In reply to #2

Re: Torque to force conversion.

11/17/2006 3:04 AM

"It's important that we all use the same units as it minimizes the risks of misunderstandings like the one that caused the loss of a Mars explorer space craft."

I bet JPL's project manager went ballistic when the code review turned up that anomalous furlongs per fortnight rate multiplier.

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#4
In reply to #3

Re: Torque to force conversion.

11/17/2006 6:46 AM

Yet another example of a SNAFU caused by international standards for measurements being FUBAR and a failure by all the programmers to RTFM.

I bet the project manager use every expletive he could think of and probably came up with a few new ones when he found out what happened. I can just see the two programmers in his office pointing at each other saying it was his fault then blaming it on some poor schmuk that wasn't even there at the time. It reminds me of the 9 stages of any engineering project

1/. Rejoice and at the project getting the go ahead and have an all day management only lunch.

2/. Analyze the task at hand with a committee of unqualified managers who's first decision is that they should be meeting at a tropical 5 star resort.

3/. Distribute of tasks to the least appropriate and qualified engineers.

4/. Throw the specifications away

5/. Build what management thinks the specifications should have said.

6/. Form another committee of unqualified managers that meet at another tropical 5 star resort to analyze what went wrong.

7/. Search for the guilty.

8/. Find and persecute the uninvolved.

9/. Blame the engineer that said it wouldn't work in the first place for letting management make the wrong decision.

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#5
In reply to #4

Re: Torque to force conversion.

11/17/2006 12:48 PM

Your 9 steps are perfect, but if it's a US Govt project, then there is a step 10.

10. Rinse-repeat

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#6
In reply to #5

Re: Torque to force conversion.

11/17/2006 4:51 PM

Amen to that....and never forget on Government Projects the following:

No good deed shall go unpunished!!!!

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#9
In reply to #2

Re: Torque to force conversion.

11/18/2006 12:01 AM

Hi Masu,

I understand that part but I would like to know anothyer aspect to that equation as to which radius I need to take into consideration.

  • The shaft(load) is on a horizontal plane and is about 25Kg or say 54 pound
  • Taking into consideration the Static Friction(μF=0.35) let us say it adds 9Kg or 20 pound and th dynamic friction is very less...
  • The motor has a driving gear of say radius X and is connected to the shaft via a belt of thickness Y, And the Shaft is fixed to the belt. Also a set of pulleys are used to provide sufficient tension on the belt and avoid sag.
  • In such a case what radius do I need to consider for conversion of torque to force, would that be X+Y(in standard units of meter) or just Xmeter or would it be a more complex function of the gears or pulleys included in the drive system.

Please do clarify my doubts.

And thank you for your advice on the uasage of units... I learnt that lesson a long time back in 6th grade while doing problems in physics(during assignments and class tests) But it was fun (and surprising) learning that mistakes like that are done even in Nasa and other examples along the message trail...

with regards and thanks,


Vishal Vijay Singh

(+919940131329)

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#12
In reply to #9

Re: Torque to force conversion.

11/18/2006 2:30 AM

I always believe in the truism that a picture is worth a thousand words so is this the sort of thing you are describing?

The driven shaft is shown in green with a radius y transfers it torque to the belt shown in grey with thickness x to pull a mass m shown in brown. The three blue pulleys are set up to even out any sudden changes in torque and to take up any slack

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#13
In reply to #12

Re: Torque to force conversion.

11/27/2006 1:28 AM

The horizontal lines represent the shaft.

There are two shafts in parallel. The combined weight is W, the time period of reciprocation is Tsec.

The belt is used to convey the torque/force to the shaft.

M - Motor

P - Pulley

I guess it is more clear now.

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#14
In reply to #12

Re: Torque to force conversion.

11/28/2006 10:55 PM

Is there a more efficient way to get it done. than using the belt and pulley??

I am thinking about using a linear stepper or likewise(given my frictional co-effs are low enough)

By now I understand the torque to force conversion. But given the load being around 25Kg(both shafts put together) and the system being belt driven, I guess I would be facing inertial problems and backlash. I don't know what I am going to do??

I just wanted to let it out that's all. Coz,I dont know if I have provided enough info here.

Ok take care. And thank you for all your help.

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#11
In reply to #2

Re: Torque to force conversion.

11/18/2006 12:07 AM

Thank you...

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#7

Re: Torque to force conversion.

11/17/2006 5:15 PM

Hi there

When rotating a shaft it is a meter of the inertia of the shaft and components rather the weight of the shaft. inertia , simply, is the product of the mass you wish to rotate and the (distance from the rotation axis)^2. There are simple formulas for different cases (geometries). If it s to complicated, try building a 3d model, and the cad software will give you the answer. Then you need to determine what is the angular acceleration (teta'') of the system, if it is a servo application, or what is the angular velocity for systems running at constant speed (teta').

The equations from here are simple.

For servo applications J*TETA''= MOMENT

for systems running at constant angular velocity teta' J*(TETA')^2=output_power

input_power=output_power/efficiency ------ to chose the motor

Efficiency depends on type and number of ballberring and geer used to get the transmission ( angular velocity).

Each transmission is usually of 2 geers and 4 beerings, and usually are greeter the 0.9 efficiency.

If your application is of direct drive from the motor with no transmission and you have 2 ballbering here the weight of the system and type of ballbeering counts. Friction coeff. Of ballbering is between 0.001 to 0.1 . Lose of moment due to ballbeering friction is Mf=fric_coeff.*weight*(ballbeering inside radii). Moment=power/ teta' in wat/rad/sec=Newton*meter

the force that creates the moment is F=(output_power/teta'+Mf)/x

Micha

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#8
In reply to #7

Re: Torque to force conversion.

11/17/2006 11:40 PM

Hello Michakiko,

Thank you for your formulas. At present I do not require it as I am not rotating the shaft but using a system to move the shaft linearly along its axis in the horizontal plane.

But thank you once again as the formula will help me in some of my other projects.

vishal...

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Anonymous Poster (1); kkrazyt85 (1); masu (3); Mevel123 (1); michakiko (1); nagaraja (1); vish_al210 (6)

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