Dimensional Analysis

Key Questions

  • How does dimensional analysis work?

    Dimensional analysis is simply a way of testing whether the base units of a given equation work out. It operates on a simple principle: the units you have on one side of an equation must match those that you have on the other.

    It's analogous to baking a cake. Your cake can only have what ingredients you put in to it.

    Consider the following equation:

    #F=ma#

    Force in base units is #kg*m/s^2#
    Mass is just #kg#
    Acceleration is #m/s^2#

    So now let's evaluate this by plugging in these units into the equation:

    #kg*m/s^2 = kg*m/s^2#

    This equation works, therefore it can be classified as dimensionally correct.

    Let's look at another equation:

    #Deltax=v_ot+1/2a^2t^2#

    #Deltax# is a distance, measured in meters (#m#)
    #v_o# is a velocity, measured in meters per second (#m/s#)
    #a# is acceleration, measured in meters per second squared (#m/s^2#)
    #t# is time, measured in seconds (#s#)

    Now let's just plug everything in:

    #m = m/s*s+(m/s^2)^2*s^2#

    Notice that I have not included the #1/2#. This is because coefficients do not matter in dimensional analysis because they don't really change the dimension (i.e. half a mass is still a mass).

    Now we simplify:

    #m = m+(m^2/s^4)s^2#

    #m = m+(m^2/s^2)#

    Since this equation does not work out, this equation is not dimensionally correct.

    Hope that helped :)

    Darshan Senthil · 1 · Dec 26 2014
  • What is dimensional analysis used for?

    Dimensional Analysis is used in engineering as a simple way to check one's work.

    After someone solves a problem, especially a conversion, they need someway to check that are correct. An easy way to do that is to check the units you were given, and seen if they make sense for what you ended up with.

    For example, if you have #13 color(white)(0) kg xx 15 color(white)(0) m/s^2# and you say that equals #195 N#

    To check your work, let's just with the units:

    #kg xx m/s^2 = N#

    You want both sides of the equation to look the same. Right niw they don't but let;s break the unit Newton #(N)# into it's base parts -- #(kg xx m)/s^2#. Now let's see:

    #kg xx m/s^2 = (kg xx m)/s^2#

    There, now they look the same! So that is an indication that we did the math right. It doesn't guarantee we're correct, but if dimensional analysis shows we made a mistake, it gives us a chance to catch it

    Rachel · 1 · Jul 1 2018

Questions

Forms of Linear Equations