Trying to do the equations for a Spring.

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In summary: The 6.18 is still listed as N, not Nm but taken that its a measurement of torque then it has to be Nm right? If a Newton meter is one Newton of force on the end of a 1m lever then if I times it by 100 do I get 618 Newton centimeters? I'm guessing that the 9.80665 is the conversion of N to Kg. So if I divide my 618 Ncm by this I get the 63 Kg cms?Right. 9.80665 meters per second squared is the standard acceleration of gravity. One kilogram force is the force it takes to support one kilogram mass in a place where Earth's gravity is equal to
  • #1
pete
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I was looking at this spring to make a constant torque motor:

https://www.ondrives.com/sr119

They come in many sizes and I've been trying to get my head around the specifications so I buy the right one.
Take the one I linked above as an example, it's listed as 6.18N (63kg) at the top of the page and further down under the specifications tab it lists it as having 63 Kg cms or 6.18 N Torque along with the dimensions of the drum the spring would wind onto.

Considering the size of the spring, this seemed to me like a very small number in N and a very large one in Kg. My maths is not too good. Neither is my physics but I spent some time looking up how this works and what the equations are so I can try and resolve what weight this spring would counterbalance.

This is the bit I’m hoping someone can look at and tell me if I’m doing this right. I tried posting this in the Engineering forum but no one answered so I thought I'd try my luck over here. I'm not looking for someone to do the work for me, I just need someone to look at it and see if I'm going in the right direction before I commit to buying things.

I took the diameter of the Torque output drum: Diameter [D3] in the specifications list, and divided it to get a Radius of 87mm, so 8.7cm. Then I changed it to meters, so .087m. Then I divided the Torque, 6.18 N by this to get roughly 71 N force. So a barrel of the same diameter as D3 with a cable wound around it will give me about 7Kg counterbalance over the extension of the spring, 12.8 meters. This still seems low to me but then I have calculated for a barrel with a 17cm diameter. Once you get down to an output shaft of around 15-20mm it would be around the 63kg point.

How did I do? Am I reading the specifications right, do I have the right equation or anything else you notice?
 
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  • #2
pete said:
They come in many sizes and I've been trying to get my head around the specifications so I buy the right one.
Take the one I linked above as an example, it's listed as 6.18N (63kg) at the top of the page and further down under the specifications tab it lists it as having 63 Kg cms or 6.18 N Torque along with the dimensions of the drum the spring would wind onto.
The only sense I can make of the specifications is if "N" really means Newton-meters. And if the Kg is kilograms force.

63 times 9.80665 divided by 100 gives 6.18 as one would expect for that interpretation.
 
  • #3
It has 6.18N (63kg) at the top of the page but in the drop-down list of specifications it has the 63Kg as cms which I presume means centimeters squared.

The 6.18 is still listed as N, not Nm but taken that its a measurement of torque then it has to be Nm right? If a Newton meter is one Newton of force on the end of a 1m lever then if I times it by 100 do I get 618 Newton centimeters? I'm guessing that the 9.80665 is the conversion of N to Kg. So if I divide my 618 Ncm by this I get the 63 Kg cms?
 
  • #4
pete said:
It has 6.18N (63kg) at the top of the page but in the drop-down list of specifications it has the 63Kg as cms which I presume means centimeters squared.
I take it as centimeters. The "s" is because it is plural.
The 6.18 is still listed as N, not Nm but taken that its a measurement of torque then it has to be Nm right? If a Newton meter is one Newton of force on the end of a 1m lever then if I times it by 100 do I get 618 Newton centimeters? I'm guessing that the 9.80665 is the conversion of N to Kg. So if I divide my 618 Ncm by this I get the 63 Kg cms?
Right. 9.80665 meters per second squared is the standard acceleration of gravity. One kilogram force is the force it takes to support one kilogram mass in a place where Earth's gravity is equal to that standard.

One kilogram force = 9.80665 Newtons.
 
  • #5
Sorry had to go to bed. It was getting very late here.

1 Kg is x 9.80665. I thought they were all derived units so I presumed 1N would be 1Kg. I know now, I looked it up, acceleration not gravity.
So x 9.80665, very neat.

Not squared, I was a bit confused how something that looked like a pressure number could add up to torque. Kg cms Like Newton meters, do you write Nms for the plural of Newton meters too, 10 Nms? Looks a bit odd but then what would I know.

So if a Nm is one Newton applied one meter along a lever. 63Kg cms must be a 63Kg force one cm along a lever? If this is right then the specifications make much more sense. 63Kg cms means you can see at a glance roughly what the strength of the spring is in relation to the diameter of an output barrel and the weight you need to counterbalance.

Travel is a bit confusing to me. I would like it in revolutions of the output shaft but it's given as a length of the spring. I have the dimensions of the barrel it should be wound on and off but then as the spring winds on or off it the OD would change each unit of travel. Sounds complicated.

I only need a rough estimate of this so I just used the diameter of the one marked as the Torque Output Drum.

Am I understanding this right now?
Thank you by the way for taking the time to help me with this.
 
  • #6
pete said:
Not squared, I was a bit confused how something that looked like a pressure number could add up to torque. Kg cms Like Newton meters, do you write Nms for the plural of Newton meters too, 10 Nms? Looks a bit odd but then what would I know.
In a physics classroom, one would avoid using "s" to indicate a plural. It would be way to easy to confuse it for an "s" meaning seconds. You could do it if you were spelling all the units out in full but not with abbreviations.

So you could write "10 Nm" or "10 Newton-meters".

[When written out in full, the "Newton" is not capitalized. When abbreviated as "N", it is capitalized. This is important so that one can distinguish between Nm (Newton-meter) and nm (nanometer)]

In a physics classroom one would never use kg as a unit of force. Nor would one capitalize the K. The abbreviation for kilogram is "kg", not "Kg".
So if a Nm is one Newton applied one meter along a lever. 63Kg cms must be a 63Kg force one cm along a lever?
Yes, you've got it.
Travel is a bit confusing to me. I would like it in revolutions of the output shaft but it's given as a length of the spring. I have the dimensions of the barrel it should be wound on and off but then as the spring winds on or off it the OD would change each unit of travel. Sounds complicated.
In real life I am just a network engineer. A guy who configures routers and designs networks and who paid attention in physics class many years ago. I do not know standard practice for how mechanical engineers quote spring specifications in catalogs. The state of the practice seems deplorable.
 
  • #7
When you're starting with as little as me. It's these small things that get taken for granted in descriptions and online information and keep tripping me up. I can see how the torque equation works now so it'll stick in my head. Once I know how things work they always do.

I don't know if you looked at the link but these springs are made to be built into a constant torque motor. You have the spring wound onto one spindle then feed it onto a spindle next to it. By winding it from one spindle to the next you get a constant torque on the output drum's spindle over however many revolutions it takes to wind it from the first drum to the second. This one is a strip of sprung 301 stainless steel, 50mm wide, 0.5mm thick and 12,8 meters long. Its a beautiful bit of kit really but because of this, I think the specifications are tailored to this use. Compression springs are listed by this company differently with a Load in N and Spring Rate in N/mm. I guess I deal with that when I get to it.

I'm still unclear on how to derive the distance but I suspect that it's obvious enough if I knew how to look.

I might give the supplier a call but I learned years ago when I first started this that if you call and they get the impression you haven't got a clue, and I don't have a clue, they can become a lot less helpful. I understand where they are coming from, a private person trying to buy one or two of something is not really worth much to them so I try to have all my ducks in a row before I get in contact. Being on good terms with these guys goes a long way.
 

Related to Trying to do the equations for a Spring.

1. What is the equation for the spring constant?

The equation for the spring constant is k = F/x, where k is the spring constant, F is the force applied to the spring, and x is the displacement of the spring.

2. How do you calculate the potential energy of a spring?

The potential energy of a spring is calculated using the equation PE = 1/2 * k * x^2, where PE is the potential energy, k is the spring constant, and x is the displacement of the spring.

3. What is Hooke's Law and how does it relate to spring equations?

Hooke's Law states that the force applied to a spring is directly proportional to the spring's displacement. This means that as the spring is stretched or compressed, the force applied to it increases or decreases proportionally. This law is used to derive the equations for springs.

4. How do you incorporate damping into spring equations?

Damping is a force that opposes the motion of a spring and can cause it to lose energy. To incorporate damping into spring equations, a damping coefficient is added to the equation for the spring constant, resulting in k = F/x - c * v, where c is the damping coefficient and v is the velocity of the spring.

5. Can spring equations be used for all types of springs?

Yes, spring equations can be used for all types of springs, including linear springs, torsion springs, and compression springs. However, the specific equations and constants used may vary depending on the type of spring and its characteristics.

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