# What exactly is a kilogram-meter?

This question isn't just about kilogram-meters, it's about multiplied units in general.

I have a good mental conception of divided units, e.g. meters per second or grams per cubic meter. Meters per second just (usually) means how many meters are traversed by an object each second. Simple.

But I don't really have a mental conception of what it means when units are multiplied, e.g. a kilogram-meter. Can anyone explain in simple physical terms what a kilogram-meter is?

• – Steve Byrnes Jul 15 '12 at 20:09

A kilogram-meter doesn't have an intuitative physical significance. Laar's product examples are meaningful for composite units like force or volts, but when dealing with products of basic units like kg, m, s, moles, amperes, physical meaning seems to only occur when there's a divisor like seconds, per your comments. So for example, kilogram-meter without a second divisor (for momentum) is intuitively meaningless.

• @dirtside I wouldn't be so quick to accept this negative answer - someone else might come along with the insight you need to make sense of it. – Nathaniel Jul 15 '12 at 20:41
• Couldn't a square-meter or a foot*inch be considered a combined measure? also isn't foot-pound a fairly common measure of pressure desnity over an area (Which would be the same as kilogram-meter)? This doesn't seem like a good answer. – Bill K Jul 16 '12 at 16:05
• The more I think about it the worse this answer gets--Multiplied units are exactly as useful as divided. ManHours, footpounds, KWH (Killowatt Hours)... I can't downvote, but it really should be deleted so as not to confuse future knowledge seekers. – Bill K Jul 16 '12 at 16:18
• How is Foot-Pound or Killowat-hours not a product of the basic units (or meter-meter for that matter)? Also, you state your answer as a fact, not your opinion. Do opinions even enter into it? Sorry, I am not intending to attack you, just suggesting that this is misleading--perhaps you could replace your answer with one more like @in_wolfram_we_trust's answer? – Bill K Jul 16 '12 at 19:57
• Bill: the pound in foot-pound is a unit of force, and the kilowatt in kilowatt-hour is a unit of power. They are composite not basic units. We'll just have to disagree on this topic. – Michael Luciuk Jul 17 '12 at 1:26

The kilogram-metre (or more often the kilogram-kilometre) is a unit of measurement indicating that 1 kg (lets assume of coal) has been moved 1 km (towards, for example, a power station). It's used by big freight companies, governments etc. as one metric of how much hauling they're doing. As you can imagine they're often they're very concerned with CO$_{2}$ emissions per kilogram-metre.

(I realise you asked for an explanation of general multiplied units, rather than about the kg-m specifically, but there are already a few general answers).

• I agree with that. – Emilio Pisanty Dec 10 '12 at 1:02
• It is also used quite often as tightening torque for screws and bolts, making the same mistakes done when confusing mass and weight. – DarioP May 28 '14 at 11:24

By your question you could see divided units as rate, an amount of one quantity would be changed based on the amount of another. When looking it at that way you could think about multiplied units as conserved quantities, if you would double one you should half the other to have the same effect. Some interesting examples could be:

• Power $P = U I$ thus in volt ampere;
• Torque $\tau = F r$ thus in Newton meters (rotating a door).

Combining these two idea's about divided and multiplied units could be done by division on one of the sides, taking the identity $U = RI$ or in units $$\mathrm{V} = \Omega \mathrm{A} \Leftrightarrow \Omega = \frac{\mathrm{V}}{\mathrm{A}}.$$ Where you could take two views. From the point of division the resistance is the amount of power required at a certain current. Or from the point of multiplication, you could keep the same voltage difference by increasing the value of the resistor by the same factor as decreasing the current.

Multiplied units often turn up in "double proportion" situations, where a particular quantity is proportional to multiple dependent variables. To take a deliberately non-physics example, a difficult task might need either many people to work on it or people to work on it for a long time, so its difficulty is proportional to both the number of workers and the time spent and could thus be measured in "person-days". A six person-day job would require two people to work for three days to complete it, or three for two days, or whatever. (The validity or otherwise of the model in which these two working schemes are equivalent is left as an exercise for the reader!)

Moving to a fairly exact parallel from a physics context, an electron-volt - or, to be pedantic, an electron-charge-volt - is a good unit of energy because the work done to move a charge through a potential difference is proportional to both the charge and the potential difference. The same goes for a coulomb-volt, although of course we just call that a joule.

Sure, you can explain this by cancelling out SI base units, but to some extent that's a red herring. If we defined the coulomb and the volt as base units, we would be perfectly happy to define our unit of energy as a coulomb volt.

• See, "person-day" (or man-hour or whatever) makes intuitive sense to me -- the amount of work done by one person in one hour. Kilogram-meter is... what? The amount of energy needed to move one kilogram one meter? – dirtside Jul 16 '12 at 0:13
• @dirtside A kilogram meter could come up in balancing, where it would be the 'torque'-like equivalent of one kilogram at one meter from the turning point. I say 'torque'-like because torque would be measured in Newton-meter, which only differs by the gravitational accelerating $g$ as extra multiplier. – Laar Jul 16 '12 at 15:27
• @dirtside Well... yes, actually, in the sense that if all you ever wanted to do was move a mass vertically in a uniform gravitational field, then you could indeed measure the work you did in kilogram-metres. Of course you need to include the gravitational acceleration with its proper units to compare to any other form of energy. But my general point is that this is useful wherever something is both proportional to mass and distance(/displacement). In this case it's energy, in in_wolfram_we_trust's excellent example it's emitted CO2. With some imagination it could be almost anything! – Ant Jul 17 '12 at 11:59

Many composite units are more intuitively understood when reformulated using other units. An example is N*s being the unit of momentum. This does follow clearly from F = dp/dt, but if you rephrase it as 1 N*s = 1 (kg*m/s²)*s = 1 kg*m/s, it's intuitively clear from p = m*v.

Similarly, and Rob Maimon already pointed this out, 1 kg*m = 1 (kg*m²) / m. Now kg*m² of course is the unit of moment of inertia (dI = r² dm), which makes kg*m a unit of moment of inertia per unit length.

When confronted with a non-obvious combination of units, you can try to rewrite it until it makes sense, but it's really always best to just look at the context, and everything should be obvious rather quickly.

Consider a collection of a large number of small spinning spheres lying on a line, which are all spun up and spun down using electromagnetic fields. A kg-m is the unit of the moment of inertia per unit length on this line.

Consider a car moving on a road with streetlamps. The faster the streetlamps go by, the more momentum the car has. The momentum per unit freqency of the car has units kg-m. A kg-m is the unit of the momentum per hz.

• Why is this a bad answer? It's a physical quantity of dimensions kg-m. – Ron Maimon Jul 15 '12 at 22:25

The unit $\textrm{kg m}$ can also be seen as $\textrm{kg}$ per wavenumber: $\textrm{kg m}=\textrm{kg}/\textrm{m}^{-1}$, so that if $M(x)$ is a function of position with units of mass - say, the mass on a rod to the left of some point - then its Fourier transform, $$\hat{M}(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty M(x)e^{ikx}dx,$$ has units of $\textrm{kg m}$, or $\textrm{kg}$ per wavenumber.

This is arguably a bit contrived, but it fits the bill. It can be adapted by changing $\textrm{kg}$ to some other useful dimension but the quantities one might want to Fourier-transform are usually some kind of density or other.

However, there is a close analogue which you do find in practice, if you change meters to seconds. In the laser cooling of atoms, for example, one usually worries about the energy delivered to the atoms, in the form of the power per unit area, which is pretty standard. However, since the transitions involved are pretty narrow, it is also important to have one's lasers very sharply in tune with the transition, since the atoms will not respond to frequency components outside the transition bandwidth. Thus, the relevant quantity is something in watts-per-square-meter per megahertz, i.e. times seconds.

• Maybe I'm just new here, but I was looking for a simple, intuitive explanation. This involves integrals, which gave me headaches even back when I did know how they worked, back in college. :) – dirtside Jul 16 '12 at 3:49
• I know it's complicated and relatively contrived, but it's the simplest I can think of. If I think of something simpler I'll certainly post it here, but I doubt there will be anything elementary. – Emilio Pisanty Jul 16 '12 at 8:54

Emilio was pretty much onto it with the integration. I’d like to expand on that.

Consider an object. It has some number of kilograms of mass, and it has moved some number of meters of distance. The product of these two numbers (in $kg\:m$) is the sum of the object’s momentum at every instant during transit. As if you were tracking its momentum and keeping a running total.

Such a continuous sum is just an integral.

You can take a sum and divide it by a count to get an average; when you divide by the number of seconds that the object was travelling, you get its average momentum for that period. Your result is in $\frac{kg\:m}{s}$, which is the familiar unit of momentum.

A Joule is a newton-meter, that is, $\frac{kg\:m^2}{s^2}$. The combined units in the numerator $kg\:m^2$ make you think “mass-area”, but in reality the kilograms, the metres, and the other metres are all separate quantities. Divide them by seconds, and you get angular momentum, or action. Divide again, you get force. Divide yet again, and you get force onset rate. Or separate force into a product of momentum $\frac{kg\:m}{s}$ and velocity $\frac{m}{s}$.

Since multiplication is commutative and associative with units, there are often many equally valid intuitive models for a given combination of units. Go with what works for you or the problem at hand, and always be ready to see things another way.

I have a torque wrench which reads in kg•M, but it actually displays kilogram-force • meters. The kg-force is the weight of one kg in one gee. Thus, 9.8 N•M = 1 kgf• M.

Using kg as force is not SI, but is rather common.

A $kg \, m$ may not have much intuitive explanation but a $N m$ certainly does: it might be seen as the energy require to lift something weighing $1\, N$ by $1 \, m$ and if you had something heavier or were lifting a greater distance then the energy required would be proportionally greater, depending on the product of the resistant force and the distance (and if the force changes then the integral of the force over the distance).

kg.m is the imbalance of an object, relative to its axis of rotation. To balance it, you've got to add a counterweight on the opposite side of the same kg.m magnitude. i.e 8x3 on the one side counteracts 6x4 on the other side - to keep it in equilibrium. (Think engine flywheel...)