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. *******************************************************
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

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)
--
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You've slipped a decimal point. Divide by 10. Failing that you're heading for
designing the Hubble telescope which was a similar silly cock up in the first
place as I understand it.
Dave Baker - Puma Race Engines

formatting link

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.

"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

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

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

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. *******************************************************
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

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. *******************************************************
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

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

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

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

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

formatting link

(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
Google happliy returns:
(((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.
I hope Google's common sense approach to interpretation of unit conversions
both answers your original question and terminates the original thread.
Hth,
Fred Klingener

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