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The conversion in SI units is 1 ksi = 6.895 MPa, or 1 MPa = 0.145 ksi. The megapound per square inch (Mpsi) is another multiple equal to a million psi. It is used in mechanics for the elastic modulus of materials, especially for metals. [5] The conversion in SI units is 1 Mpsi = 6.895 GPa, or 1 GPa = 0.145 Mpsi.
By definition, 36 AWG is 0.005 inches in diameter, and 0000 AWG is 0.46 inches in diameter. The ratio of these diameters is 1:92, and there are 40 gauge sizes from 36 to 0000, or 39 steps. Because each successive gauge number increases cross sectional area by a constant multiple, diameters vary geometrically.
The weight per unit length diminishes by an average of approximately 20% at each step. Because the weight per unit length is related to the cross sectional area, and therefore to the square of the diameter, the diameter diminishes by approximately 10.6%:
No. 7/0, the largest size, is 0.50 inches (500 mils or 12.7 mm) in diameter (250 000 circular mils in cross-sectional area), and the smallest, No. 50, is 0.001 inches (1 mil or 25.4 μm) in diameter (1 circular mil [cross-sectional area] or 0.7854 millionths of a square inch).
Fabric "weight" is often specified as mass per unit area, grams per square meter (gsm) or ounces per square yard. It is also sometimes specified in ounces per yard in a standard width for the particular cloth. One gram per square meter equals 0.0295 ounces per square yard; one ounce per square yard equals 33.9 grams per square meter.
When using "hand in" to convert to hands and inches, the rounded hands and inches values are equivalent, and use the same fraction, if any. Special rounding of the inches value only occurs when "hand in" is the output.
The older United States Standard Gauge is based upon 40 lb per square foot per inch thick. Gauge is defined differently for ferrous (iron-based) and non-ferrous metals (e.g. aluminium and brass). The gauge thicknesses shown in column 2 (U.S. standard sheet and plate iron and steel decimal inch (mm)) seem somewhat arbitrary.
Load bearing is required to carry (the weight which is being exerted through the combined weights of the shaft and any other direct weights on the shaft and measured in pounds-force per square inch): Formula: L = W / (I.D × L.O.B.). Example: Determine the load on a bearing of a 2-inch I.D. bearing, 5 inches long and carrying a weight of 3,100 lbf: