If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Intro to dimensional analysis

Sal shows how we can treat units of measurement algebraically, and use these tools in order to convert between different units of the same quantity. Created by Sal Khan.

Want to join the conversation?

  • blobby green style avatar for user elise
    In the practice, many of the problems have the problems expressed in meters squared or cubed, but the video does not explain how to handle the numbers when converting from say, cm3 to m3 (sorry I don't know how to subscript!) Are there any videos doing this type of rate conversion?
    (21 votes)
    Default Khan Academy avatar avatar for user
  • mr pants teal style avatar for user NavNalajala
    At ,i don't understand how he does the hour/second formula. It always gets me and I don't understand how it works. Can anybody help please?
    (7 votes)
    Default Khan Academy avatar avatar for user
  • leaf grey style avatar for user Solipse
    @, Sal calls for multiplying mh/s by something that has hours as the denominator. I don't quite understand why that is.
    (5 votes)
    Default Khan Academy avatar avatar for user
    • duskpin tree style avatar for user Bian Lee
      He is doing that to get rid of "hour", and to replace it with "seconds". By making "hours" the denominator, the "hours" will cancel out since (hour)/(hour) is 1, and then the only time unit left is "seconds". When he is making "hours" the denominator, he also has to make the numerator 3600 "seconds" to keep the value same as before, since (3600 sec)/1h = 1 and multiplying any number (except 0) by 1 will always be the number you multiplied to, meaning it wouldn't change the value. What Sal is teaching us is how we can change the unit while keeping the value same
      (13 votes)
  • piceratops sapling style avatar for user Hedayat
    I'm doing this in my chemistry class. The teacher does it in a very complicated way but the video has it in an algebraic way and not a chemistry way. I'm confused. Whats the difference?
    (7 votes)
    Default Khan Academy avatar avatar for user
  • starky sapling style avatar for user medisha02
    Would this work using any formula, like a=F/m? Like if I have a force acting on an object of 15 N and a the mass of the object as 58 kg, would I be able to figure out the acceleration using dimensional analysis?
    (7 votes)
    Default Khan Academy avatar avatar for user
  • piceratops ultimate style avatar for user Ashley O'brien
    I'm having trouble with the process of conversion, I'm having trouble understanding the process used here. Judged on the practice, there feels like there is more to it than this.
    (7 votes)
    Default Khan Academy avatar avatar for user
  • boggle yellow style avatar for user andy
    Hello in one practice question it says:
    The batting Wang uses to fill quilts has a thermal conductivity rate of 0.03 watts (W) per meter (m) per degree Celsius (C). and the equation I am given is

    W / (m * C)

    why is it meter * by celsius?

    I thought it would be
    W / (m / C)
    since it is meter per degree Celsius...

    Thank you!
    (5 votes)
    Default Khan Academy avatar avatar for user
  • starky tree style avatar for user Ani-Jay
    Does anyone know a better way of explaining what he's talking about? I am having difficulties applying what he said in the videos to the practice problems he's giving me.
    (2 votes)
    Default Khan Academy avatar avatar for user
    • piceratops ultimate style avatar for user Nolan Ryzen Terrence
      There is nothing much to worry We know distance = Speed * Time
      We know the units for each of them
      Distance = metre
      Speed = metre/second
      Time = Second
      Now Using the equation:
      D = S * T
      D(m) = S(m/s) * T(s)
      The quantity in the bracket is their unit


      Let's say I am going at 20 m/s speed in 20 seconds. What is the distance I have traveled?
      >>
      D(m) = 20 m/s * 20 s
      20*20=> 400 (That seems so simple isn't it?)
      Lastly about the units:
      The s in the denominator(speed) and s in the numerator(seconds) cancel out. You are left with the meter on RHS which is the same unit on LHS, this is the basis of DIMENSIONAL (Using Dimensions) ANALYSIS (I don't have to give the meaning do I?) or DA


      Let's say I travel at 5m/s for 1 hour. What is the distance?
      If you are pretty fast your mind will think of converting time to s
      Conversion:
      1 hour has 3600 seconds(Ok mind is steady)
      1hr(given in question) * 3600 s/1hr (read as 3600seconds per hour, logically that is correct right?)
      Now calc the numbers and the units of hours cancel out leaving 3600 seconds.
      Multiply with speed
      5 m / s * 3600 s The seconds Unit cancels.You are left with 18000m.
      If you have understood till here then you can try using DA to find 18000m in km..

      Let me know if it still confuses you.
      Nolan :)
      (13 votes)
  • starky sapling style avatar for user malcolmsheridan
    What if it doesn't say how many seconds like, "Uche pumps gasoline at a rate of 18 ."
    (4 votes)
    Default Khan Academy avatar avatar for user
  • male robot hal style avatar for user JaredAFord
    I'm still kind of confused. Could someone explain a little more to me please? Thank you.
    (5 votes)
    Default Khan Academy avatar avatar for user

Video transcript

Voiceover:We've seen multiple times in our life that distance can be viewed as rate times time. What I want to do in this video is use this fairly simple formula right over here, this fairly simple equation, to understand that units can really be viewed as algebraic objects, that you can treat them like variables as we work through a formula or an equation, which could be really, really helpful to make sure that we're getting the results in units that actually make sense. For example, if someone were to give you a rate, if they were to say a rate of, let's say, 5 meters per second, and they were to give you a time, a time of 10 seconds, then we can pretty, in a pretty straightforward way, apply this formula. We say, well, distance is equal to our rate, 5 meters per second times our time, times our time, which is 10 seconds. What's neat here is we can treat the units, as I've just said, like algebraic constructs, kind of like variables, so this would be equal to, well, multiplication, it doesn't matter what order we multiply in, so we can change the order. This is the same thing as 5 times 10, 5 times 10 times meters per second, times meters per second times seconds, times seconds. If we were to treat our units as these algebraic objects, we could say, hey, look, we have seconds divided by seconds, or you're going to have seconds in the denominator multiplied by seconds in the numerator. Those are going to cancel out, and 5 times 10, of course, is, 5 times 10, of course, is 50. We would be left with 50, and the units that we're left with are the meters, 50 meters. That's pretty neat. The units worked out. When we treated the units out like algebraic objects, they worked out so that our end units for distance were in meters, which is a unit of distance. Now you're saying, "OK, that's cute and everything, "but this seems like a little bit of too much overhead "to worry about when I'm just doing "a simple formula like this." But what I want to show you is that even with a simple formula like distance is equal to rate times time, what I just did could actually be quite useful, and this thing that I'm doing is actually called dimensional analysis. It's useful for something as simple as distance equals rate times time, but as you go into physics and chemistry and engineering, you'll see much, much, much more, I would say, hairy formulas. When you do the dimensional analysis, it makes sure that the math is working out right. It makes sure that you're getting the right units. But even with this, let's try a slightly more complicated example. Let's say that our rate is, let's say, let's keep our rate at 5 meters per second, but let's say that someone gave us the time. Instead of giving it in seconds, they give it in hours, so they say the time is equal to 1 hour. Now let's try to apply this formula. We're going to get distance is equal to 5 meters per second, 5 meters per second times time, which is 1 hour, times 1 hour. What's that going to give us? The 5 times the 1, so we multiply the 5 times the 1, that's just going to give us 5. But then remember, we have to treat the units algebraically. We're going to do our dimensional analysis, so it's 5, so we have meters per second times hours, times hours, or you could say 5 meter hours per second. Well, this doesn't look like a ... This isn't a set of units that we know that makes sense to us. This doesn't feel like our traditional units of distance, so we want to cancel this out in some way. It might jump out of you, well, if we can get rid of this hours, if we can express it in terms of seconds, then that would cancel here, and we'd be left with just the meters, which is a unit of distance that we're familiar with. So how do we do that? We'd want to multiply this thing by something that has hours in the denominator and seconds in the numerator, times essentially seconds per hour. How many seconds are there per hour? Well, there are 3,600 ... Let me do this in a ... I'll do it in this color. There are 3,600 seconds per hour, or you could even say that there are 3,600 seconds for every 1 hour. Now when you multiply, these hours will cancel with these hours, these seconds will cancel with those seconds, and we are left with, we are left with 5 times 3,600. What is that? That's 5 times 3,000 would be 15,000, 5 times 600 is another 3,000, so that is equal to 18,000. The only units that we're left with, we just have the meters there. 18- Oh, it's 18,000, 18,000, 18,000 meters. We're done. We've now expressed our distance in terms of units that we recognize. If you go 5 meters per second for 1 hour, you will go 18,000 meters. But let's just use our little dimensional analysis muscles a little bit more. What if we didn't want the answer in meters but we wanted the answer in kilometers? What could we do? Well, we could take that 18,000 meters, 18,000 meters, and if we could multiply it by something that has meters in the denominator, meters in the denominator and kilometers in the numerator, then these meters would cancel out, and we'd be left with the kilometers. So what can we multiply it so we're not really changing the value? We want to multiply it by essentially 1, so we want to write equivalent things in the numerator and the denominator. So 1 kilometer is equivalent to, equivalent to 1,000 meters. One way to think about it, we're just multiplying this thing by 1, 1 kilometer over 1,000 meters. Well, 1 kilometer is 1,000 meters, so this thing is equivalent to 1. But what's neat is when you multiply, we have meters canceling with meters, and so you're left with 18,000 divided by 1,000 is equal to 18. And then the only units we're left with is the kilometers, and we are done. We have re-expressed our distance instead of in meters in terms of kilometers.