I bumped into your site while trying to understand why the US were still stuck on the Imperial system.

I am Swiss, working as a consultant engineer in Nigeria. When doing the conversion to metric, here, they kept the Imperial at hand. Result: you buy steel sheets of 4' x 8' x 3mm... and the measuring tapes have both sides, cm and inches. Your carpenter will supply cabinets one meter wide by two feet deep, the fuel pumps sell liters of fuel, while at the black market you buy the fuel by the gallon... One gets used to it!

Now, we use to import machinery from the US, and I'm into aviation as well. So, of course, I've been wondering for a while: why on Earth do the US stick to the Imperial?

During my trips in the States, I realized that people there simply do not know what the metric system is, and only think of it as just another complicated set of meaningless units and conversion tables.

I wrote the following some time ago, for a post on the net.

"As an engineer who learned the metric system in primary school, having to deal with American engineering and its (pardon the expression, but I'll make my point further down) inane measuring system is a constant source of puzzlement.

How on earth can anyone devise a working mechanism, using a system based on fractions, where nothing corresponds to anything? No wonder that the countries using the Imperial system tend to stick with one engineering design for centuries: once you've got all your measurements together, you certainly don't want to start it over again!

Discussing with American engineers, I realized that they simply do not understand the underlying implication of the metric system. Used to memorize tables and gauges, and to just know that so-and-so size will fit so-and-so purpose, they are totally unwilling to even think about learning everything from scratch again.

I once proposed a simple test to a colleague of mine in Dallas. Here it is:

Take an aluminum fuel tank. Given the dimensions in feet and inches,
calculate:

- The volume in gallons
- The volume in fluid ounces
- The weight of the fuel in pounds at full tank
- The weight of the empty tank in pounds

Do this without referring to any table, and without an engineering calculator. You may use a simple "4 operations" calculator, though... we are not trying to test your math skills.

Time the whole process.

These are the data (any figure would do, actually):

- Width, 2'
- Depth, 2 ½'
- Height, 6"
- Thickness 0.04"
- Aluminum specific density 2.7
- Fuel specific density 0.7.

The answer I got from my friend was simple: "But I don't think I can do it without my tables...". After a considerable amount of time and fumbling and grumbling, he came out with results that, once checked against tables, turned out to be wild guesses. Not close enough to do anything meaningful with them.

I then sat down, asked him to give me random figures for the dimensions, and set to work in metric. Off-head. Yes, I cheated a bit on the last item, and used a calculator for the empty tank weight. My time was 1 minute 17 seconds. Without a sweat, working at a leisurely pace.

How comes? Well, just look at this... the Imperial System in all its
majesty:

1 Gallon (liquid) = 231 Cu. In
1 Gallon (dry) = 268.8 Cu. In.
1 Fluid Ounce = 1.805 Cu. In
1 Pint = 16 Fl. Oz.
1 Quart = 2 Pints
1 Gallon = 4 Quarts
1 Firkin = 9 Gallons
1 Cubic Foot = 1728 Cu. In
1 Foot = 12 Inches
1 Yard = 3 Feet
1 Mile = 1760 Yards

We can continue like this all day long.

So, to solve our little problem above, we first have to remember that 1 Cu.
In. of water weighs 0.578 ounces or thereabout (but you do remember it, don't you?).

We are not here to split hairs on conversions, just to make a reasonably correct calculation, usable in every day's life.

Thus we have:

2 x 2.5 x 0.5 = 2.5 Cu. Ft (Volume of the tank)

2.5 x 1728 = 4320 Cu. In (Volume in Cu. In.)

4320 / 231 = 18.7 Gallons (First answer)

18.7 x 128 = 2393.6 Fl. Oz (Second answer - I'm sure you know that 1 Gallon
(liquid) is 128 Fl. Oz., and 231 Cu. In....)

4320 x 0.578 = 2496.96 Oz water weight equivalent (water being the reference for specific density)

2496.96 / 16 x 0.7 = 109.242 Lb of fuel (Third answer)

Here, I'd like to point out that one Troy pound contains 12 Troy ounces, whereas one Avoirdupois pound contains 16 of its own particular brand of ounces... I had to scratch off some and redo my calculations after re-reading the tables.

(2 x 2.5 x 2) + (2 x 0.5 x 2) + (0.5 x 2.5 x 2) = 14.5 sq. ft. x 144 = 2088 sq. in. (Developed area of the tank)

2088 x 0.04 = 83.52 Cu. In. (Volume of the tank's material)

83.52 x 0.578 x 2.7 = 130.34 Oz of aluminum, which, once divided by 16, gives us 8.14 lbs for the empty tank. Fourth and last answer... Pheew!

One thing I can tell you, is that this exercise certainly took me more than
1 minute 17", and that I extensively used a calculator, consulted tables, swore, and fumbled. I am not even sure of my results, did not cross-check them, and wouldn't be able to cross-check them at a glance.

Let's see our metric problem, now...

- Width, 40 cm
- Depth, 50 cm
- Height, 15 cm
- Thickness 0.8 mm
- Aluminum specific gravity 2.7
- Fuel specific gravity 0.7.

And with that, give:

- The volume in liters
- The volume in cubic centimeters
- The weight of the fuel in kilograms at full tank
- The weight of the empty tank in kilograms

Well, here we go.

4 x 5 x 1.5 = 30 lt (First answer)

30,000 cm3 (Second answer)

30 x 0.7 = 21 Kg (Third answer)

(0.4 x 0.5 x 2) + (0.4 x 0.15 x 2) + (0.15 x 0.5 x 2) = 0.67 m2 (Developed area of the tank)

0.67 x 2.7 x 0.8 = 1.447 Kg. Last answer (Yes, I used the calculator on this one, didn't want to melt my brain). You can stop the chrono.

I did not need tables to remember that 1 dm3 = 1 lt, and 10 cm = 1 dm.
Therefore, 4 x 5 x 1.5 gives me liters straight away.

Again, I knew off-head that 1 lt = 1000 cm3. And calculating 30 x 0.7 without a calculator is not great feat, once one knows that 1 lt of water is
1 kg...

As you can see, I'm jumping from one unit to another just by moving the decimal. I used m2 for the last calculation, since 1mm over 1 m2 gives you 1 kg of standard density... and I can get my weight/thickness/area ratio directly.

No intellectual gymnastic, and I've never been very good at calculus, make no mistakes.

The basics of the metric system are easily summed up:

Length units: Meter, multiplied or divide by 10, 100, 1000 as is convenient Area units: Square meter, multiplied or divided by 10, etc.
Volume units: Cubic meter, multiplied etc..
Liquid units: Liter, etc.
Weight units: Gram, etc
Now, please note the points that really make the Imperial system look like a tool for the suppression of all engineering:

- One liter is a cube of 10 cm x 10 cm x 10 cm.

- One kilogram is equivalent to one liter of water at standard temperature.

- Water is used as standard density material (value 1)

Seamless integration of length, area, volume, and weight. In my place, if you can't master this stuff by age 9, you are good for a specialized institution (or you become a politician)."

Go to Measurement Standards (SI in the U.S.)