"Do the math!"- Sega 16-bit video game-set commercial, c. mid-1980's

Units in physics and chemistry have an interesting property in that they can be treated as algebraic variables which you can multiply and divide and use for grouping. This allows you to perform conversions accurately and with confidence. With this knowledge, you can even derive your own conversion factors easily.Converting units is such a common task in physics and chemistry that there are several videos on the subject on YouTube.

One property of the various physics constants is that "they make the units come out right" (as one of my teachers told us). For example, the units of the Gravitational Constant in the MKS system (meter-kilogram-second) are (kg

^{–1}m^{3}s^{-2}). When you calculate the gravitational force between two bodies excluding that constant, the units come out as (kg^{2}m^{-2}). When you include the constant, the units multiply together thus:(kgBy definition, the MKS unit of force, the Newton, is 1 kg m s^{–1}m^{3}s^{-2}) (kg^{2}m^{-2}) → (kg^{(2-1)}m^{(3-2)}s^{-2}) → (kg m s^{-2})^{-2}, so the constant's units straightened out those of the calculation to leave us with newtons. QEF (Quod erat faciendum, "that which had to be done").This is typically one of the first subjects covered in chemistry and physics and is essential knowledge. This page will demonstrate the technique and also how to derive your own conversion factors.

Transitioning from high school to college was an exciting time in my life during which I learned a lot of really cool things and which basically turned me into a life-long learner.For example, in junior college I took a "physical science" class which was basically algebraic physics, physics for non-science majors but still one step above "Physics for Poets" (the actual course title for an actual course). Later I took the first semester of real physics, which is based on calculus, such that it is said that Newton had to invent calculus in order to express and work with the ideas he was discovering and working with for physics. There are some subjects in physics, such as the moment of inertia used in angular momentum, which are simple to express in calculus but which require long tables of formulae in algebra (eg, a different formula for the moment of inertia of each different shape of rotating body). In such cases, the algebra just allows you to calculate the value for that specific case, whereas the calculus expresses the concepts.

In many of the lectures in this class, the instructor would start with a few formulae and describe what they tell us of the relationships between its various factors (eg, directly proportional, inversely proportional, inverse-square law). Then he would solve for one of the factors in one formula and substitute that solution for a factor in another formula in order to come up with a new formula and new relationships. Wow! I mean, it has become common practice since then, but when it was first demonstrated to us ... wow!

But one real mind-blower he showed us (in a manner which seemed like an obvious aside at the time) was that you can treat the units of measurement for each factor as algebraic variables in themselves and thus end up with the correct units in your solution. Yes, that is correct; you can treat units the same as algebraic variables. That was one of the cooler things I learned in that "physical science" class. And then he pointed out that one of the more important functions of constants in physics is to make the units come out right -- I will demonstrate that below with the Constant of Gravitation ( G = 6.674×10

^{-11}m^{3}×kg^{-1}×s^{-2}) as it applies in Newton's Law of Universal Gravitation.Outside of the obvious coolness factor, a very practical use for this fact is in deriving conversion factors. There are many times when we need to convert from one unit to another and we cannot always find the conversion factor(s) we need in a handy reference book. Remember, that was decades before the Internet, so having the right reference book was essential (now we can't even find handy reference books anymore except in used book stores). And many times, the conversion we need to perform is too unusual for it to even appear in any reference table. Or else we need to work through a chain of conversions (eg, converting miles down to inches). And maybe also have to account for exponents (eg, converting cubic inches to cubic yards). In such cases, we end up needing to create those conversion factors ourselves and we need to be able to keep everything straight.

Later at my first job as a software engineer I quickly became the go-to guy for conversion factors even though I kept trying to teach them how to do it themselves. Now mind you, these were exotic units (eg, binary angular measurement (BAMs)) that had been defined specifically for our assembly program, so you could not find them in any reference books (nor could you find them on-line since in 1983 there was no such thing as "on-line").

More recently, I have used these techniques on a couple of my web pages: DWise1: Kent Hovind's Solar Mass Loss Claim and DWise1's Creation/Evolution Page: Earth's Rotation is Slowing. In fact, I created this page in order to replace a rather lengthy footnote in the solar-mass-loss page where I derive the rate at which the sun is losing mass due to hydrogen fusion. In the page on the earth's rotation, I cite several sources but all their rates are given in different units, so I have to convert them all into one common set of units in order to compare them meaningfully. On the latter page, I rely on these conversion techniques to ensure that those units get converted correctly and accurately.

Newton's Law of Universal Gravitation is perhaps the classic example for how a constant's units will straighten out the units in the calculation.First, we must choose our system of units. In physics, those are most commonly:

There are a number of humorous systems of units, including the infamous furlong-firkin-fortnight (FFF) system. For this example, we will choose the large-scale metric system, MKS.

- Large-scale metric: meters, kilograms, seconds (MKS)
- Small-scale metric: centimeters, grams, seconds (cgs)
- US Conventional: foot-pound-second (fps)
Within each system we derive many other units based on those basic three units. Here, we need to derive the units for force. Since force is mass (kg) times acceleration and acceleration is distance divided by time squared (m/s

^{2}), the units for the unit of force, newtons (N) in MKS, are:1 N = 1 kg m/sBTW, in the cgs system the unit of force is the dyne, which is defined as 1 g cm s^{2}

= 1 kg m s^{-2}^{-2}Now to define the variables and their units in MKS:

F is the force between the masses in newtons (1 N = 1 kg m sPlease note the low magnitude of the gravitational constant. Of the four forces (gravitational, electro-magnetic, strong nuclear, weak nuclear), gravity is the weakest as is indicated by the gravitational constant. The only reason gravity seems so strong to us is because of how massive the earth (1.19×10^{-2})

G is the gravitational constant, 6.674×10^{-11}m^{3}kg^{-1}s^{-2}

m_{1}is the first mass in kilograms (kg)

m_{2}is the second mass in kilograms (kg)

r is the distance between the centers of the masses in meters squaredd (m^{2}).^{x}kg) and the sun (1.19×10^{x}kg) are.The formula for gravitational force is:

This gives us the following units calculations (using the times sign to make it explicit):G×m_{1}×m_{2}F = ---------- r^{2}kg m sm^{3}kg × kg m × kg kg m^{2}m × kg ---------- × --------- → -------- × ----- × ---- → -------- → kg m s^{-2}kg × s^{2}m^{2}s^{2}kg m^{2}s^{2}^{-2}is the definition for Newtons. Hence, the Gravitational Constant, G, has served its secondary purpose of making the units come out right.QEF

We know what the gravitational constant is in the MKS system, but what is it in the cgs system? Do we need to Google for that? Or can we just figure it out ourselves? Yeah, Goggle'ing would be what we'd usually do, but figuring it out is more fun.First, we have G = 6.674×10

^{-11}m^{3}kg^{-1}s^{-2}in MKS. To convert that to cgs, we need to change the meters to centimeters and the kilograms to grams and substitute those values in. Or as my high school algebra teacher would say, we just plug them in and grind it out:So what we come up with as the value of the gravitational constant in the cgs system is 6.674×101 m = 100 cm 1 kg = 1000 g G = 6.674×10^{-11}m^{3}kg^{-1}s^{-2}= 6.674×10^{-11}(100 cm)^{3}(1000 g)^{-1}s^{-2}= 6.674×10^{-11}× 100^{3}cm^{3}× 1000^{-1}g^{-1}s^{-2}= 6.674×10^{-11}× (10^{2})^{3}cm^{3}× (10^{3})^{-1}g^{-1}s^{-2}= 6.674×10^{-11}× 10^{6}cm^{3}× 10^{-3}g^{-1}s^{-2}= 6.674×10^{-11}× 10^{6}× 10^{-3}cm^{3}g^{-1}s^{-2}= 6.674×10^{(-11+6-3)}cm^{3}g^{-1}s^{-2}= 6.674×10^{(-8)}cm^{3}g^{-1}s^{-2}^{(-8)}cm^{3}g^{-1}s^{-2}. Did we get it right?Looking it up in Wikipedia, we find it's equal to about 6.674×10

^{-8}cm^{3}g^{-1}s^{-2}. So we nailed it. And we also have gone through an example where the terms being substituted have exponents.

Convert the density of the sun's core from g/cm^{3}to kg/m^{3}:1 kg = 1000 gQEF

1 m = 100 cm

Density of Sun's core = 162.2 g/cm^{3}1 kg 1 (100 cm)^{3}= 162.2 × (g × -------) × (---- × -----------) 1000 g cm^{3}1 m^{3}g 1 kg 1 100^{3}cm^{3}= 162.2 × ( --- × -----) × ( ---- × ----------) g 1000 cm^{3}1 m^{3}1 kg cm^{3}10^{6}= 162.2 × ( -----) × ( ------ × -------) 10^{3}cm^{3}1 m^{3}1 kg 10^{6}= 162.2 × ----- × ------- 10^{3}1 m^{3}10^{6}1 kg = 162.2 × ----- × ------- 10^{3}1 m^{3}= 162.2 × (10^{3}kg/m^{3}) = 162,200 kg/m^{3}

Convert 7 days to seconds:Strategy: repeatedly multiply the value being converted by 1, which does not change its value. Of course, 1 comes in a great many different forms.Or, if you will be doing this more than once, you could have created a conversion factor for converting from days to seconds:

QED24 hours day 7 days × --------- = 7 × 24 hours × ---- = 168 hours day day 60 minutes hour 168 hours × ----------- = 168 × 60 minutes × ----- = 10,080 minutes hour hour 60 seconds minute 10,080 minutes × ----------- = 10,080 × 60 seconds × ------- = 604,800 seconds minute minute Ergo: 7 days = 604,800 secondsConversion factor for converting from days to seconds: multiply by 86,40024 hours 60 minutes 60 seconds 1 day = 1 day × -------- × ----------- × ----------- day hour minute day hour minute = 1 × 24 × 60 × 60 × ----- × ------ × -------- × seconds day hour minute 1 day = 86,400 secondsQED

As I said above, at my first job as a software engineer I was the go-to guy for conversion factors even though I kept trying to teach my co-workers how to do it themselves. Now mind you, these were exotic units (eg, binary angular measurement (BAM)) that you could not find in most reference books (nor could you find them on-line since in 1983 there was no such thing as "on-line").The procedure is simple: you just set up an equation for the same value using different units and then solve for the unit you want to convert from; eg:

Derive the conversion factor for radians to degrees, given that a full 360° circle is 2π radians:2π radians = 360°Conversion factor for converting from radians to degrees: multiply by 180/π

1 radian = (360 / 2π)°

1 radian = (180 / π)°

QED

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*First uploaded on 2019 August 13.
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