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46 RICHARDSON'S BLUEBOOK OF PROJECTION
Calculating Resistances
In practice one orders a rheostat to deliver approximately the amperage required or H.^^^St of amperes is desired for which there is no stock rheostat avaihbk he may order two of different capacities and connect them h/series or parallel In that way almost anv amoerage capacity may be built up. .
(577lt?easy to calculate any resistance required in ohms Suppose we have a 110 volt supply and propose usurg 60 amperes of current. Ohm's law tells us that Stage divided by amperes equals ohms We know a 60 amp. arc operates at approximately 55 volts hence 110 _ 55 (representing voltage drop of arc) . W, ana
^ 60 = 916 + ohms required in rheostat. In other words, first subtract the voltage drop across the .arc from the line voltage and divide the remainder by ■** de^red amperage. Result will be the amount of resistance required, expressed in ohms.
Heat
A resistance coil or grid will function under any degree of neat tl Swi 1 not fuse it. (58) However, excessive temperatures cause rapid deterioration^a^lowe^ length of life (59) Moreover, overloading resistance elements may at any time set up heat sufficient to fuse the metal and thus stop all current flow. .
60) The n\aximUm permissible resistance coil or grid temperature is 900° Fahr. This temperature will make the metal just visible in a dark room.
A dull red in daylight is approximately 1 300 degrees Fahr. At such a temperature resistance metal wdl not last long Even 900 Fahr. is too high. Five hundred decrees Fahr is as high as resistance really should be dpStedFto operate. Such a limit f^^ J^JJ the life of the coils or grids. (61) In piactice ii me metalls visible in a dark room the = amperage should be cut down until the metal is invisible in total darkness. This can be done easily if it is an adjustable rheostat If Jis nXSen you must either install an additional coil or