396 MESSRS. W. E. AYRTON AND H. KILGOUR ON THE 
diameter is about 2 ohms at 300° C. Hence the resistance of 17 centims. of wire 
0'75 mil., or 0'0191 millim. in diameter, will be about 
/ 0-152 \» 
2 (from/ 5 or 128-0 ohms - 
If, therefore, x be the required current density, the rate of production of heat in 
calories (gramme-centigrade) per second will be 
r /0'00i9i\ 3 i 3 
X 7T X ( —) [ x 128-0 X 0*239, 
which must equal the number of calories lost per second, viz., 
17 X 7T X 0-00191 X 0-02258 X 285, or 0-6546, 
therefore, x = 51330 amperes per square centimetre. If the wire, instead of being 
platinum, had been made of copper, and if the emissivity for a copper w r ire, 0"75 mil in 
diameter, were the same as that of a platinum wire of the same diameter, the current 
density that would be required to keep the copper wire at a temperature of 300° C., 
when the enclosure was at 15° C., would be about 
51330/57, 
that is, 122.600 amperes per square centimetre, or 790,500 amperes per square inch. 
[It may be useful to give the general formula which we have arrived at for 
calculating the current A amperes, required to be sent through any long thin 
platinum wire cl mils in diameter to maintain it at a temperature of 300° C. when 
the enclosure (having about the dimensions shown in fig. 1) is maintained at a 
temperature of 15° C. 
From the various measurements given in the paper it follows that the resistance 
per cubic mil of platinum at 300° C. is about 0"0086 ohm. Hence, equating the rate 
of production and loss of heat per mil length of the wire, we have 
A 2 x 0-0086 x 4 x R24 
7rd 2 
“/ (0-001135 + 0-016081 rf" 1 ), 
therefore, 
A = 1-483/(0-001135 d s + 0-016084 cP) about. 
Also, if the wire having the conditions described above be l inches long, the watts 
that must be expended on it equal approximately (0"3869 + 0"02732 cl) l. 
