CONVERSION OF UNITS IN AVALANCHE WORKA paper on basic math for converting units. If you're into converting numbers...you'll want to click to: Centre for Innovation in Mathematics Teaching It has the most extensive listing of conversion math we've been able to find. General Conversion Technique There are many ways to convert from one system of units to the next. One of the simplest methods is cross multiplication to cancel out the undesired unit. Three examples of increasing difficulty illustrate this technique. The reader is urged to work the given examples; a pocket calculator will be helpful. Example 1. Convert 24 inches to feet. Solution: (24 in) (1 ft/12 in) = 2 ft Note that multiplication is by a fraction equal to one. The numerator and denominator are chosen to cancel out the undesired unit. Example 2. Convert 10 miles per hour to feet per second. Solution: (10 mi/hr) (5280 ft/1 mi) (1 hr/3600 sec) = 14.7 ft/sec Example 3. Convert 100 ounces per square inch to pounds per square feet. Solution: (100 oz/in) (1 lb/16 oz) (12 in/1 ft) (12 in/1 ft) = 900 lbs/ft2 The Metric System There is a worldwide trend toward expressing all measurements in one system of units, the metric system. Two basic metric units commonly used in avalanche work are the meter and the kilogram, which are respectively units of distance and mass. These two basic units are the building blocks for more complex units such as speed, density, and stress. Distance A meter (m) is a little more than a yard, more precisely 3.28 feet or 39.37 inches. Meters are used for measuring mountain elevations, thickness of slab fracture, runout distance of avalanches, artillery range, etc. Problem 1. The range of a 105 mm HE round fired from a howitzer is 11,300 m. Convert this to miles. (Problem solutions appear at end of Appendix A.) Large distances are measured in kilometers (km) which is 103 meters, and small distances in millimeters (mm) which is 10-3 meters. Problem 2. How many inches in diameter is a 75 millimeter projectile? It is sometimes convenient to express certain measurements such as the amount of new snow in centimeters (cm). The centimeter is 10-2m, which is about 2.54 inches. Problem 3. John Doe wants to sell his old 7-ft skis. What size should he advertise? We all want to minimize confusion by using the least number of different units. It is argued that all our purposes are served if we stick to units based exclusively on factors of 1,000: e.g., megameters (106m), kilometers (103m), meters (1 m), milimeters (10-3m), and micrometers (10-6m). Hence, centimeters (lO2m) is not considered a preferred description of distance. However, it is tolerated on an interim basis. Units Based on Distance and Time Units of area and volume are based on distance units to second and third powers respectively (e.g., m2, m3). Problem 4. A standard density sample tube has a volume of 500 cm3. What is its volume in cubic inches? Units of speed are based on distance divided by time. Avalanche speed, for example, is measured in meters/second (m/s). A brisk walking pace is about 2 m/s. Problem 5. An avalanche is clocked at 50 m/s. What is its speed in miles per hour? Problem 6. An avalanche is triggered by a sustained precipitation intensity of 1 mm/hour. What is this in inches per hour? Mass and Density The amount of material, mass, is based on the fundamental unit, kilogram, which is slightly over 2 lbs (2.20 lbs). Density is mass per unit volume, and is properly expressed as kg/m3. The density of water is 1,000 kg/m3. Problem 7. The average density of a slab is measured as 200 kg/m3. What is its density in pounds per cubic foot? Snow density is sometimes expressed as grams per cubic centimeter. In this system, the density of water has the convenient reference of 1.0 g/cm3. Problem 8. Show that 1,000 kg/m3 equals 1 g/cm3. Force and Stress The preferred unit of force is the newton (N). The preferred unit of stress (force per unit area) is N/m2. Unfortunately, there is much confusion over force units. Although it will not be possible in this short appendix to resolve the confusion, the reader should understand that some of the confusion is due to the fact that force measuring devices such as spring scales are calibrated in units of mass (kilograms, grams), rather than newtons. Thus, it is common practice (although poor practice) to speak of avalanche impact pressures, shear stress, ram strength, etc., in units of mass (e.g., kg/m2 for pressure). Problem 9. Use the equation on page CR to calculate the shear stress of a 2 meter slab of density 250 kg/m3 on a 38 degree slope (sm 38 degrees = 0.616). (Editor note cross reference to section 3.6.) Problem 10. Use equation on page CR to calculate the impact pressure of an avalanche with speed 50 m/s, and flowing density of 100 kg/m3. From section 4.6. Although the newton (N) may not be familiar to many readers, it is handy in force and stress computations. For example, the impact force equation on page CR 4.6 is simply Impact Pressure = PV2 where the impact pressure is now in N/m2, if density is in kg/m3, and V is in m/s. Note that the newton is equivalent to (kg) (m/s2), a result that is derived from Newton's second law of motion: Force = (mass) (acceleration) Problem 11. Use the data of problem 10 to compute impact pressure in n/m2. The common unit of atmospheric pressure is the millibar (mb) which is 102 N/m2. One bar (1000 mb) equals 29.5 inches of mercury. Problem 12. What is the atmospheric pressure in inches of mercury at the 750 mb level? Temperature Snow and avalanche observations are based on the Celsius temperature scale. A comfortable room temperature is slightly over +20 degrees C. Snow melts at 0 degrees C, and is cold enough to squeak at -20 degrees C. The formulas for converting from Celsius (C) to Fahrenheit (F) are: F = 9/5 C + 32 C = 5/9 (F - 32) Problem 13. Convert -20 degrees C to degrees F. Solutions to Problems Problem 1. (11,300 m) (3.28 ft/1 m) (1 mi/5280 ft) = 7.02 mi Problem 2. (75 mm) (1m/1000 mm) (39.37 in/1 m) = 2.95 in Problem 3. (7 ft) (12 in/1 ft) (2.54 cm/1 in) = 213.4 cm Problem 4. (500 cm3) (1 in/2.54 cm) (1 in/2.54 cm)(1 in/2.54 cm)= 30.5 in3 Problem 5. (50 in/s) (3600 s/1 hr) (3.28 ft/1 m) (1 mi/5280 ft) = 112 miles/hour Problem 6. (1 mm/hr) (1 cm/10 mm) (1 in/2.54 cm) = 0.039 inches/hour Problem 7. (200 kg/m3) (2.2 lbs/1 kg) (1 m/3.28 ft)3 = 12.5 lbs/ft3 Problem 8. (1000 kg/m3) (1000 g/1 kg) (1 m/100 cm)3 = 1 g/cm3 Problem 9. Shear stress = P D sin 0 = (250 kg/m3) (2 m) (0.616) = 308 kg/m2 Problem 10. Impact pressure = PV2/g kg/m2 = 100 kg/m3 (50 m/s) (50 m/s) = 25,510 kg/m2 9.8 m2/s2 Problem 11. Impact pressure = PV2 N/m2 = 100 kg/m3 (50 m/s) (50 m/s) = 25 * 104 N/m2 Problem 12. (750 mb) (1 bar/1000 mb) (29.5 inches of Hg/1 bar) = 22.1 inches of Hg. Problem 13. F = 9/5 (-20) + 32 = -4 degrees Table 1.--Length Table 2.--Area Table 3. --volume Table 4.--Weight Table 5.--Weight per unit area Table 6.--Velocity Table 7.--Grade percent and equivalent degree of slope Table 8.-- Degree of slope ond equivalent grade percent Table 9.--Natural trigonometric functions, by half degrees Fahrenheit Celcius Temperature alinement chart forconverting Fahrenheit to celcius