Unit Conversions and Gc? Please resolve this conflict!

Factual Questions
1

1a. Let’s say you have a formula that is supposed to yield a force. Using SI units, density (kg/ft^3) is one term plugged into a formula…and, along with the other units, the units reduce to kg-m/s^2. You say "Ah-ha! That’s a Newton, so I don’t need to convert kg to Newtons. So, did you just get lucky that Gc is not needed? Yet, your conscience tells you 1kgmass = 1kgforce <> 1 N, so do you only apply Gc at the end when the units don’t work out right?

1b. Then, you run the same exercise in English units and wind up with units of lb-ft/s^2. You know that’s a weird set of English unit so you divide by Gc which (a) does yield the converted answer from (1a) above, BUT (b) the units reduce to being unitless! What’s up?

That dang Gc thing always messes me up…every dang time.

2

I don’t get you.
Are you saying that kg/ft^3 is an SI unit?
When you say “1kgmass = 1kgforce <> 1 N”, what does the “<>” mean? Obviously not “less than greater than”, so do you mean “is interchangeable with” or are you representing conversions back and forth? Is the argument that “<>” takes just the “kgforce” or the statement “1 kgmass = 1 kgforce”?
I don’t like using kilograms to describe forces and am not sure what the conventions are, but would guess that by “kgforce” we mean the weight of a kg mass, which would not be 1 N but rather about 10 N.
Sorry, but what is “Gc”? Is it the gravitational constant multiplied by some c? Is it just the gravitational constant, specifically the one relating mass and force in Earth’s gravity field?
I think it is best to do everything in SI units, and the first operation in using anything that is not already in SI is to convert it there. If you want to use English units too, use slugs.
There is too much going on here, but without the formula you’re dealing with, for me to figure out what is up. Can you post it?

3

Since there are no replies just yet…It has been a long time since I’ve studied physics but it sounds like you have a pretty basic question. However, I’m trying to figure out just what your question is. Let me see if we can clarify it so that I or someone more knowledgeable can answer it.

1a. Using SI units, density (kg/ft^3)

Well, ft are not SI units so you’ve got me there. If you have volume in ft you need to convert to m[sup]3[/sup] first (maybe you do it further down the road, but you should do those conversions first).

…the units reduce to kg-m/s^2. You say "Ah-ha! That’s a Newton, so I don’t need to convert kg to Newtons.

Well, yes, those units are Newtons, but…a kg in SI is a unit of mass, not force. You can’t convert mass to force, unless you want to assume the force of Earth’s gravity at a point on the surface…

  • So, did you just get lucky that Gc is not needed? Yet, your conscience tells you 1kgmass = 1kgforce <> 1 N, so do you only apply Gc at the end when the units don’t work out right?*

…which would be g (the force of gravity at sea level, equal to 9.80665 m/s[sup]2[/sup]), although you’re discussing Gc–do you mean the gravitational constant, usually just G (I’m very rusty on that one, though). And again, there is no such thing as 1 kg of force.

By 1b I am completely lost.

You start with an expression that is supposed to yield a force, and it does. So not sure why you are uncomfortable with that.

If you had an expression that yielded mass, and you wanted to find out the force of gravity on that mass, then you would use g.

It would also help immensely if you explained just what it is you’re trying to figure out in real-world terms before jumping down into units and equations.

4

Many programming languages use “<>” to mean “not equal to.” I assumed that here.

5

I expect it’s to do with the issues mentioned under “Use of pound as a unit of force” in the wikipedia article on Pound force.

Apart from that I’ve nothing to add, except to repeat that there’s no such unit as kg-force, and I wish they’d teach that to the people who translate Discovery programs into Norwegian.

6

This is due to the difference in the definition of N and lbf:

1 N == force required to accelerate 1 kg to 1 m s^-2 == 1 kg m s^-2
1 lbf == force exerted on 1 lbm due to gravity on Earth == 32.174… lbm ft s^-2

Suppose you come up with the quantity 42 kg m s^-2 after your calculations. You need the answer in N. Since 1 N == 1 kg m s^-2, 42 kg m s^-2 * 1 (N)/(kg m s^-2) == 42 N.

However, suppose you need to convert 42 lbm ft s^-2 to lbf. 1 lbf == 32.174… lbm ft s^-2. Therefore, 42 lbm ft s^-2 * (1 lbf / 32.174… lbm ft s^-2) == 1.3054 lbf.

Hope this helps.

7

If I understand you correctly, then you have to keep in mind that you only need that conversion when gravity actually enters your calculations. You can have both kg and N and a meaningful relationship between them without dealing with gravity. If we just look at the definition of a N then we can see one simple scenario. If we have a mass on 1 kg and we want to accelerate it by 1 m/s within 1 s then we need a force of 1 N. It doesn’t matter whether we do that floating in space or on an idealized frictionless ice surface on earth (or anywhere else.)

That was/is a reasonably common non-SI unit although it is evil.

8

Gc is an abomination used by engineers who insist on doing things like using completely different units that happen to have the same name. If you’re working in a sane unit system, such as SI, then it’s easy to know when to use Gc, because the answer is “never”. If you’re doing a calculation in SI that’s supposed to yield a force, and your answer has any units other than Newtons, then either your formula is wrong, or you made a mistake in your calculations.

9

>1kgmass = 1kgforce <> 1 N

CookingWithGas, good idea, maybe it “means not equal”. It’s tarnished, though, by the obvious inequality that preceeds it, written as an equality.

Chronos, about the abomination “Gc” - is it supposed to mean something? What? Better, is it supposed to mean two subtly different and inconvertable things at once? Or, maybe, it’s supposed to mean “moment”, which is an engineering phrase meaning “don’t ask” or “no meaning assigned” or “everybody now argue over what to say”?

Jinx, did you just make the whole thing up from random phrases so you could laugh at our attempts?

10

I’m glad to see I wasn’t the only person baffled by the OP.

And for Chronos, what exactly is this Gc of which you write? I accept your position that it’s a kludge, but what’s it (supposedly) do?

ETA: Napier’s latest wasn’t here when I loaded this page, so I wasn’t trying to pile on.

11

It’s a symptom of unit systems which use units with the same name for both force and mass, such as “pound” in (some versions of) the American system. This makes it appear that one can “convert” between force units and mass units, in the same way one might convert between, say, inches and feet. The pseudo-conversion factor used for this is Gc (the c stands for “conversion”), which is basically just the acceleration due to gravity at the Earth’s surface, except that since it’s used as a conversion factor, you pretend that it’s dimensionless like any true conversion factor.

12

I found an example of Gc online. Abomination indeed. But first, to the OP…

Jinx: I would highly encourage you to throw Gc out of your life entirely. Here’s what you should be doing for something like F=ma, regardless of the units.

Ex 1: m=3 kg, a=9.8 m/s[sup]2[/sup].
Finding F…
F=(3 kg)*(9.8 m/s[sup]2[/sup])
F=29.4 kg m / s[sup]2[/sup]
But, a newton is defined as kg m / s[sup]2[/sup], so we’re there:
F=29.4 N

Ex 2: m=3 kg, a=32.2 ft/s[sup]2[/sup].
Finding F…
F=(3 kg)*(32 ft/s[sup]2[/sup])
F=96.6 kg ft / s[sup]2[/sup]
Now we’ve got a weird combo unit for our answer. The solution: multiply or divide by 1. In particular, you can multiply or divide by any ratio of the form A/B as long as A and B are equal (since A/B would thus be 1). Here, we note that
1 m = 3.281 ft
Thus,
1 = (1 m/3.281 ft)
Multiplying by 1:
F=96.6 kg ft / s[sup]2[/sup] * (1)
F=96.6 kg ft / s[sup]2[/sup] * (1 m / 3.281 ft)
The “ft” in the numerator cancels the “ft” in the denominator. Notice that I’m viewing the units as actual factors multiplied into the expression, not as separate pieces off to the side.
F=(96.6/3.281) kg m / s[sup]2[/sup]
F=(29.4) kg m / s[sup]2[/sup]
We again have kg m / s[sup]2[/sup], which we can replace with the equivalent N:
F=29.4 N

We get the same answer, even though in the first case someone decided to report the acceleration in ft/s[sup]2[/sup]. One final example:

Ex 3: m=6.6 lb, a=32.2 ft/s[sup]2[/sup].
Finding F…
F=(6.6 lb)*(32 ft/s[sup]2[/sup])
F=212.5 lb ft / s[sup]2[/sup]
Here we will need to multiply by 1 twice if we want to get to N. Our two “ones”:
1 = (1 m/3.281 ft)
1 = (1 kg/2.20 lb)
Multiplying by 1:
F=212.5 lb ft / s[sup]2[/sup] * (1 m / 3.281 ft) * (1 kg / 2.20 lb)
The "ft"s cancel and the "lb"s cancel, leaving
F=(212.5/3.281/2.20) kg m / s[sup]2[/sup]
F=(29.4) kg m / s[sup]2[/sup]
F=29.4 N

The beautiful thing about this approach is that it can detect many common errors. That is, if you plug in a number for acceleration that is not an acceleration (say, a velocity), you’ll never be able to convert it to newtons. The Gc thing is just taking appropriate multiply-by-one factors for various input unit choices and putting them in a table. But, as the saying goes, teach a man to fish and he eats for a lifetime. The Gc tables will limit what you can calculate comfortably.

Regrading Gc: What I’ve found online differs slightly from Chronos’s description. ISTM that Gc is used more generally for unit conversions, with a different value for different expected input units. Thus F=ma/G[sub]c[/sub], and if you will be working with mass in pounds and acceleration in ft/s[sup]2[/sup] and you want force in N, you’d look up G[sub]c[/sub]=7.22. (It pains me not to give that number units.) I didn’t dig too deep, though, because in either case, I’m going to back away slowly and pretend like no one is being taught this.

13

This.

The crux of the misunderstanding is that the fundamental units of SI are mass, length, and time; the fundamental units of the imperial system are force, length, and time. All the units for dynamic situations can be derived from these three fundamental units, but the imperial system makes it a pain in the ass.

14

  1. Dang! :smack: I had a MAJOR blooper in the OP…sorry! :smack: The start of the OP should have read “kg/m^3”! Sorry all!

  2. Yes, it is an accurate statement put in math short-hand that 1 kgmass = 1 kgforce <> 1 Newton. An expression can have more than one equal sign, ya know. (Like, a= b =c.) In my case, one is simply an inequality, that’s all.

  3. Chronos, Santo Rugger, and Pasta offer some good advice. Yes, Gc is because engineers have bastardized the units for us. Perhaps the root of the problem stems from trying to Americanize the metric system? I mean, cans of soup express weight in kg. No such animal! It’s like a unicorn of science.
    (Whew! Good thing the unit “slugs” need not appear on a can of soup!) So, someone created the rule that 1 kg-mass = 1 kg-force…removing the need for Newtons. (But, they went on to say 1 lb-force = 1 lb-mass! Criminal!)

About Gc, though: Chronos claims “you pretend Gc is dimensionless…”. I wasn’t taught it like that. I was taught Gc had the same units as gravity. Often, you see G/Gc tossed into an equation which reduces to “1”. This just adds to the confusion over its purpose. (What the heck is the point of G/Gc, anyway?).

4a. Santo Rugger offers good advice on a subtle point. The basic English units are FORCE, distance, and time; however, science is more concerned with mass most likely. Hence, you divide the English result by gravity. This will help me keep things straight.

4b. Last, 4a explains the last part of my OP…if we assume Gc is dimensionless as Chronos suggests. [I hate to say it, but I really got f’d by profs who didn’t give a damn about teaching.]

Thanks all for the group effort!

15

I agree it IS confusing, but those in the sciences need to know that “bastardized units” exist. And, like anything else, when taught with a simple example or two, it is quite clear how to master. (After reading through this thread of replies, I now understand why, in your example above, Gc has a different value than expected.)

16

You know, I’d never thought about how crappy it must be to be an engineer in an “Imperial Units” world before.

Now I know. Gc, indeed. :suicide:

17

Only BASIC-derived ones. A wider variety of commonly-used languages (the ones not called BASIC or Basic) use !=.

18

My advice to the OP: Use different units for mass and force. kg should always be mass, so that 1 kg m/s[sup]2[/sup] = 1 N. If for some god awful reason someone is giving you a force in kg-force units, just immediately rewrite this in Newtons. (1 kg-force = 9.8 kg m/s[sup]2[/sup] = 9.8 N)

If you are dealing with both pound-mass and pound-force, I would denote them differently, say lbm vs. lbf. That way, at least you know whether a given number written on your paper corresponds to a mass or a force.
Where Gc comes in is when someone specifies a mass by telling you the force that mass feels at standard gravity. So, if I say a mass of one pound, you can view this as
1 lbm = 0.45 kg (in which case you don’t include Gc at all),
or equivalently you can view it as
1 lbm = 1 lbf / Gc = 1 lbf / (32 ft/s[sup]2[/sup])

If you want to express the acceleration in m/s[sup]2[/sup]
1 lbm = 1 lbf / Gc = 1 lbf / (9.8 m/s[sup]2[/sup])

If you want to express the force in terms of Newtons, use
1 lbm = 1 lbf / Gc = 4.45 N / Gc = 4.45 N / 32 ft/s[sup]2[/sup] = 1 N / [(32/4.45) ft/s[sup]2[/sup]] = 1 lbf / Gc(new)

Here we changed the value of Gc to absorb the conversion factor. Of course, if we already know the individual conversion factors there’s no need to absorb them all into a single factor. We can just apply any conversions one at a time, and I think that’s the less error prone way to go. The only reason to put them all in a single factor is so you can just look them all up at once, but as Pasta points out it’s easier to check your work if you convert your units explicitly instead of just looking up one factor that changes them all at once.

Using just one big conversion factor also makes it seem like you need Gc to change your units. But you don’t. You can always just multiply by the equivalent of 1, like Pasta said. The reason you have Gc with units of acceleration is because you converted from a mass unit to a force unit. There was no need to pull those other numerical factors into Gc except to let you look up the combined conversion factor in a book.

Note in the above examples you always want to make sure you’re only converting force to force, mass to mass, etc. Which is why you don’t want to use the same symbol for force units and mass units.

19

U.S. engineer here…

The whole concept of g[sub]c[/sub] arises because in the USCS (U.S. Customary System), both force and mass are expressed in pounds. (The USCS does not use slugs as a unit of mass, as was done in the imperial system from which it arose.) The pound-mass (lbm) is defined as that mass that exerts a force of 1 pound (lbf) at the standard value of gravity at the Earth’s surface (g = 32.174 ft/s[sup]2[/sup]). (Note that g[sub]c[/sub] is a universal conversion factor that should not be confused with the local acceleration of gravity g, which may vary from the standard value of 32.174 ft/s[sup]2[/sup].)

Therefore, the pound-force (lbf) is that force which accelerates one pound-mass (lbm) at 32.174 ft/s[sup]2[/sup]. Thus, 1 lbf = 32.174 lbm-ft/s[sup]2[/sup]. The expression 32.174 lbm-ft/(lbf-s[sup]2[/sup]) is designated as g[sub]c[/sub] and is used to resolve expressions involving both mass and force expressed as pounds. For instance, in writing Newton’s second law, the equation would be written as F = ma/g[sub]c[/sub], where F is in lbf, m in lbm, and a is in ft/s[sup]2[/sup].

I agree with Chronos that the concept of g[sub]c[/sub] is indeed an abomination, particularly with more complex calculations involving viscosity, etc. Many times I have found it easier to convert all of my units to SI, do my calculations, and convert the answer back to USCS. Other times I find myself working in a mixture of USCS and SI units. When working out calculations, it is even more critical for U.S. engineers to always write out the units, or you are sure to screw up the calcs.

20

Suddenly, the whole “Mars lander crashed because we were using different units” makes more sense.

Which brings me to my question… why on earth are you still using imperial for engineering? As you say, you convert into SI to perform the calculations anyway, and it’s not like the hoi polloi need to know what units you’re working in. Is it really harder to pour concrete in meters? I’m sure China does metric like the rest of the world. So why stay in imperial?