# Spring forces

• posted

During a previoue foraty into this territory, I was ked to the following set of calculations

d = wire diam (inches)

D = Coil mean Dia (inches)

N = No of free turns

G = 11.5 * 10^6

k (expressed in pounds per inch) =(d^4 * G)/(8*N*D^3)

For a

1" wirediam 8" diam spring 10 free turn

spring, this calc gives 280.76 lb/inch

or for my purposes, 50.24 Kg/Cm

My question:

Is that 50.24 Kg/cm actually 4922 N/cm?

I think it does, but I would like a check on this.

I hate it when somebody says "a Kg is 9.8 newton....because you weigh ten times as many newtons..because of gravity." For my purposes, gravity does not apply above.

Thanks for any help.

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Sometimes in a workplace you find snot on the wall of the toilet cubicles. You feel "What sort of twisted child would do this?"....the internet seems full of them. It's very sad

• posted

A kilogram is a mass. Newton is force (kilogram meter per second squared, to be exact; comes from the classic F = ma). For an object to have weight, that is, exert a force, there must be an applied acceleration. On Earth, most of the time that acceleration is 1G, or 9.8m/s^2. (The seconds squared means velocity (m/s) of a free-falling object goes up another 9.8m/s in 1s of fall, or the position changes 9.8^2 meters in 1s, if velocity started at zero.)

So really, whoever gave you that info, bastardized the SI units. Ironic since they were invented to PREVENT these confusions..... but yeah I think your number above is correct.

Tim (finally going to sleep, BTW mornin' Oz)

-- "I have misplaced my pants." - Homer Simpson | Electronics,

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• posted

You've slipped a decimal point. Divide by 10. Failing that you're heading for designing the Hubble telescope which was a similar silly c*ck up in the first place as I understand it.

Dave Baker - Puma Race Engines

I'm not at all sure why women like men. We're argumentative, childish, unsociable and extremely unappealing naked. I'm quite grateful they do though.

• posted

"Most of the time" should really have nothing to do with it, if this is accurately expressed in kgf/cm. If that's the case, you don't use the local acceleration of gravity, but the standard acceleration of gravity, which for this purpose (it might differ for the purpose of defining pounds force, which don't have an official definition, but we often use the same value there as well, 9.80665 m/s²).

No. It isn't bastardized SI units. It is pre-SI.

The kilogram force was a legitimate unit long before the SI was introduced in 1960. In fact, grams force were endorsed by the CGPM back in 1901, when they were made well defined units for the first time when the CGPM officially adopted a "standard acceleration of gravity" of 980.665 cm/s² for this purpose. This "standard acceleration of gravity" is not a concept of physics, it is strictly a concept of metrology. It serves no purpose other than defining units of force in terms of units of mass.

You are right, of course, that the kilogram force is not a part of SI, and not acceptable for use with SI. But until SI was introduced in

1960, kilograms force were quite legitimate units.

Of course, we still see far too many vestiges of the use of this once-legitimate unit. I have a torque wrench in "meter kilograms" and they are still readily available. We still see pressure gauges in "kg/cm²" units. Thrusts of rocket and jet engines are still expressed in kilograms force (e.g., Tom Clancy's nonfiction _Airborne_), and those were the normal units in the Russian space program until at least the late 1980s or early 1990s.

Note that if you stick to SI, a "standard" acceleration of gravity serves no purpose.

BTW, getting back to the original question,

No. While 50.24 kgf/cm would be 492.7 N/cm (different only in last digit, and assuming that you truly did have four significant digits in the original measurement whihc doesn't seem real likely in actual fact), that would be better expressed in SI as 49.27 kN/m. Avoid the prefixes which are not powers of 1000. Centimeters are good for your hat size and volumes in cubic centimeters, and little else. Even in the days of the obsolete centimeter-gram-second systems, N/cm wouldn't have been used, since then the obsolete unit of force used were dynes, not the newtons of the meter-kilogram-second systems such as SI.

Gene Nygaard

• posted

This, of course, is not a dimensionless number. For it to work in your formula below where the result is in lbf/in, the units of this constant would have to be lbf/in^2

If that's what you think you need for "your purposes," maybe it is time to rethink things. First, the unit of mass is "kg" not Kg, and the unit of length is "cm" not "Cm"--but you shouldn't actually be using either one of them.

First of all, why do "your purposes" involve a spring constant in obsolete metric units, when you have your length measurements in obsolete English units? Get rid of all of the Fred Flintstone Units, both metric and English.

For distances in millimeters, and a spring constant in N/mm, you could convert

G = 11.5 x 10^6 lbf/in^2 = 79300 N/mm^2

1 in = 25.4 mm using 1 lbf = 4.44822 N

I would imagine that the value of this G is dependent upon the material used.

Then

k = d^4 G/(8 N D^3) = (25.4 mm)^4 x 79300 N/mm^2/(8 x 10 x (203.2 mm)^3)

= 49.17529296875 N/mm

which then needs to be rounded appropriately, likely to 49 N/mm, certainly not to as many places as you carried your results, but since you were still using your figures as intermediate results that would have been okay as long as you rounded appropriately when you came to the end of your calculations. Note that no matter how precisely you measure the diameters involved, you can never have more than 2 or 3 significant digits in the final result, because that's all we have for the G used in your formula.

then you need to properly round this, depending not only on the

Make sure you include the units of G so that you can see that the units come out right in your result.

BTW, your 280.76 lbf/in would convert to 50.138 kgf/cm, not 50.24 kgf/cm, if you use the same acceleration to define both pounds force and kilograms force, so that kilograms force have the same relationship to pounds force as kilograms have to pounds (we no longer have independent standards for pounds; they are officially defined as

0.45359237 kg, exactly).

No. See the other responses.

So, gravity doesn't apply. However, if you are damn fool enough to use kilograms force (or pounds force, for that matter), then the

*standard* acceleration of gravity certainly does apply.

But don't use those obsolete units of force. Use newtons.

Gene Nygaard

• posted

I should have explained this a little better. I chose those units to use with millimeters, but of course

79 300 N/mm^2 = 79 300 000 000 N/m2

which is, of course, 79.3 GPa, since a newton per square meter is a pascal, and "G" is the symbol for the prefix giga-.

Those gigapascals are the units you are likely to see used in tables which list this modulus for various materials in SI units. So I'm just explaining how those units relate to the ones I used.

In the formula you used, if the modulus G is in pascals (convert the prefix to powers of 10), and both diameters in meters, the result will be in newtons per meter, and then you can choose an approriate prefix to use with one or the other of those units (it's generally not a good idea to put a prefix on both of them) when expressing the final result.

Gene Nygaard

• posted

On Sun, 16 May 2004 09:59:25 -0500, Gene Nygaard vaguely proposed a theory ......and in reply I say!: remove ns from my header address to reply via email

How can the force on a spring be measuerd in newtons per m2? It's a force, not a pressure.

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Sometimes in a workplace you find snot on the wall of the toilet cubicles. You feel "What sort of twisted child would do this?"....the internet seems full of them. It's very sad

• posted

On Sun, 16 May 2004 17:11:23 +0800, Old Nick vaguely proposed a theory ......and in reply I say!: remove ns from my header address to reply via email

Well so far I've had my typos corrected, had lectrures about theory and been accused of being a dam fool.

Thanks guys.

Having got over all the crap and my frustration and anger, I _amy_ dig through the enormous amnout oif extraneous stuff in Nygard's reply adn find an answer, or some of it.

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Sometimes in a workplace you find snot on the wall of the toilet cubicles. You feel "What sort of twisted child would do this?"....the internet seems full of them. It's very sad

• posted

On 16 May 2004 06:39:55 -0700, snipped-for-privacy@nccray.com (Gene Nygaard) vaguely proposed a theory ......and in reply I say!: remove ns from my header address to reply via email

Precisely. SO I get this munged formula, and the units are loosely used. I needed clarification or a better folrmula.

*******************************************************

Sometimes in a workplace you find snot on the wall of the toilet cubicles. You feel "What sort of twisted child would do this?"....the internet seems full of them. It's very sad

• posted

It's that modulus (can't remember exactly what it is called; maybe Young's modulus?) in the formula which has those units of strain (which are the same as the units of pressure).

The result of your formula is a spring constant k, at least in some direction. This has units of force divided by length; it tells you how much force it takes to move the spring a certain distance in that direction. For example, you might have a spring scale where each millimeter that you stretch it out corresponds to 10 newtons of force.

Gene Nygaard

• posted

Yikes. It's the 'shear modulus,' the ratio of shear stress (F/L^2) to shear strain (dimensionless - a distortion angle in radians). Young's modulus is the ratio of normal stress (F/L^2) to linear strain (dimensionless - L/L).

The units of strain are NOT the same as the units of stress.

Hth,

Fred Klingener

• posted

For anyone still able to stomach this wretched little thread, I'll offer the following common-sense approach to the problem: Use the Google calculator

(shear modulus in pounds/inch^2)

Plugging this:

((1 inch)^4*11.5*10^6 pounds/inch^2)/(8*10*(8 inch)^3) in pounds per inch

into the Google search block, we get

(((1 inch)^4) * 11.5 * ((10^6) (pounds / (inch^2)))) / (8 * 10 * ((8 inch)^3)) = 280.761719 pounds per inch

Cool. So far so good.

Then

Here Google assumes that you completely understand the technical, theoretical, and philosophical issues and that you choose to ignore them temporarily to get a useful practical result:

Given the following line:

((1 inch)^4*11.5*10^6 pounds/inch^2)/(8*10*(8 inch)^3) in kilograms per centimeter

(((1 inch)^4) * 11.5 * ((10^6) (pounds / (inch^2)))) / (8 * 10 * ((8 inch)^3)) = 50.138336 kilograms per centimeter

Close enough.

Finally,

Actually, no.

((1 inch)^4*11.5*10^6 pounds/inch^2)/(8*10*(8 inch)^3) in newtons per centimeter

returns:

(((1 inch)^4) * 11.5 * ((10^6) (pounds force / (inch^2)))) / (8 * 10 * ((8 inch)^3)) = 491.689113 Newtons per centimeter

Fine, except for your decimal point.