|Lastufka Labs -||Reference|
One of the most frustrating aspects of engineering is converting from one set of units to another. It can be a real pain, especially when you find out you did the conversion wrong after using your result in a complicated equation. The goal of this page is to help you get your unit conversions right the first time!
Though it is true that the international metric system has been taught in schools for years (except for metric time "chrons"), many (perhaps most) US engineering shops still use the Imperial system of units, feet, pounds, gallons, etc. Why use the Imperial units in these manuals? Mostly because the specifications in race regulations use the Imperial system.
When evaluating and discussing results obtained from mathematical models, one desires that the values work out to be small whole numbers as much as possible. Measuring raingutter regatta boats in "furlongs" would not only produce very small fractions, but would severly challenge one's "normal" ability to compare the measurements. So angstroms were invented for atomic theory, and parsecs for intergalactic travel, just to keep the number of digits under control.
The gist of the problem is scale. A good system of units used in a model should "fit" the scale of the model. Scale is handled by decimation (factors of 10) in the metric system and by more arbitrary factors in the Imperial system. In these manuals, the primary units fit to scale are radians, ounces, seconds and inches a system of measurement I call "ROSI" (pronounced rose-ee). Note, a different arrangement produces "ISOR" (pronounced eye-sore)!
|radian (rad)||There are 2 pi radians in a complete circle.|
|ounce (oz)||There are 16 ounces in an Imperial pound (lb). Though many think of this as a unit of weight, weight is really a force. So pounds and ounces are units of force. It is not correct to simply convert pounds (or ounces) to kilogrammes, a unit of mass. The force is converted to mass first by dividing by the acceleration of gravity in the appropriate units. The conversion from mass (Imperial slugs!) in the Imperial system to mass in the metric system is one of the more error-prone tasks in physical modeling. Ounces are often used as a measure of volume, the volume of water weighing 1/6 of a pound at "normal" room temperture (about 60 degrees F) and pressure (about 14.7 lb/in2).|
|second (s)||There are 60 seconds in a minute (min) and 60 minutes in an hour (hr).|
|inch (in)||There are 12 inches in a foot (ft), three feet in a yard (yd) and 5280 feet in a mile (mi).|
|CLASS||MULTIPLY||BY||TO GET||OR BY||TO GET|
Note, mass in ROSI units is obtained in practice by measuring weight in ounces, then dividing by the gravitational acceleration, g. g is about 386.088 in/s2 at sealevel on planet earth.
Note, to get values in ROSI units from the metric or Imperial systems, just divide the original value by the number indicated for the oposite (ie, inverse) conversion.
It is often common practice to convert a measure in weight to mass and vice-versa. What does it mean to do this? Suppose we have an object that weighs 1 pound (lb). How much "stuff" (mass) is in it? The Imperial unit of mass is the slug (I guess slugs are known for their inertia ;-). To get slugs from pounds one divides pounds by the acceleration of gravity in feet per second squared, 32.174 f/s2 at sea level on planet earth. Thus slugs have the primary units of lb/(in/s2) = lbs2/in. So 1 lb has 1/32.174 slugs of "stuff" in it.
Going the other way, we see that 1 slug has the force of 32.174 pounds at sea level on earth. That's a lot of "stuff" - about the same amount as a cubic foot of wood.
Applying this method to kilograms (kg), we start with newtons the metric unit of force and weight. In metric units, the acceleration of gravity is 9.80665 meters per second squared at sea level on planet earth. So 1 newton is 1/9.80665 kilograms and 1 kilogram is 9.80665 newtons.
Now, if we measure an object weighing one pound on a scale that reads in newtons, we are comparing weight with weight. The scale would read 4.44822 newtons. Suppose we compute the mass using both systems. We divide 1 lb by 32.174 ft/s2 and 4.44822 n by 9.80665 m/s. We get 1/32.174 slugs = 4.44822/9.80665 kg. Evaluating we get,
Equation A: 0.031081 slugs = 0.45359 kg
How many kilograms are in a slug? 1 slug = 0.45359/0.031081 kg = 14.59380 kg.
Note that the mass of 1 pound at sealevel on earth is 0.031081 slugs. So, from equation A above we see immediately that there are 0.45359 kg of "stuff" in 1 pound. Converting pounds directly to kilograms only works where you know the acceleration of gravity in both systems in the place where the measurement takes place. Both the 4.44822 n measurement and the acceleration of gravity in m/s2 would be different on another planet or in space so the 1 pound would contain a different fraction of kilograms, not 0.45359.
How much does 1 ozs2/in of ROSI mass weigh at sealevel on our fair planet? Force, that is weight in the present case, is mass time acceleration. That acceleration is gravity. In ROSI units it is 386.088 in/s2 at sealevel, etc.. So 1 ozs2/in * 386.088 in/s2 is 386.088 oz! Since there's 16 ounces in a pound, that's 24.1305 lb. So 1 ozs2/in of mass is pretty big. Above, we learned that 1 slug weighs 32.174 pounds. Dividing one weight by the other gives us the conversion between the masses ozs2/in and slugs. There are 24.1305 pounds of ozs2/in to every 32.174 pounds of slugs = 24.1305 to 32.174 = 3 to 4. So one ozs2/in weighs only as much as 3/4 of a slug. It would weigh 3/4 of a slug anywhere, not just on earth. So one ozs2/in is 3/4 of a slug.
|Lastufka Labs -||Reference|
|Copyright © 1998, 2002 by Michael Lastufka, All rights reserved worldwide.|