August, '11] WOODWORTH: FUMIGATION TENT LEAKAGE 379 



The time required to kill will be somewhat dependent upon the den- 

 sity of the gas. Years ago thirty minutes were employed for fumi- 

 gation but all fumigators now use forty to fifty minutes. One of the 

 investigations now under way is the determining of the killing strength 

 of gas for different times of exposure. 



We do not know when the maximum density of gas is obtained in 

 the generation which sometimes takes as much as fifteen minutes; 

 it is probable that a greater or less time elapses after the killing has been 

 accomplished before the tents are actually removed. Under average 

 conditions the present period may be considered justified. When we 

 have sufficient data we may be able to reduce the time in certain cases 

 and advise the extending of it in others. For the present, however, 

 we must accept the time and suppose that the dose of four ounces 

 for a ten-foot tree maintains the killing strength of gas for the neces- 

 sary period under the average conditions of leakage and it is not neces- 

 sary to know exactly what this period is. 



To calculate the dose for trees of clifTerent size and for tents of differ- 

 ent leakage we need but to know that the above dose applies to .25% 

 leakage for a 10 ft. tree and that in the case of other trees the ratio 

 of increase or decrease of dose corresponds approximately with the 

 area, that is, when the volumes are in the ratio of 1 : 8, the dose is in the 

 ratio of 1:4. 



The correct dose for a tent consists of two factors, (1) the theoretical 

 amount necessary to kill when the tent is tight; (2) the amount of 

 increase to allow for leakage. When one tent has twice the diameter 

 of another, provided they are of the same shape, they have the ratio 

 1:8 in volume and if 1 is the theoretical dose of the smaller, 8 is the 

 theoretical dose of the other. The second factor for these two tents 

 consists of numbers, one the square root of the other. If half the gas 

 escapes in ten minutes, half of the remainder will have escaped at the 

 end of twenty minutes. Now with a tent of twice the diameter the 

 loss in twenty minutes will only be that occurring in the smaller tent 

 in ten minutes. The dose to compensate for the losses in these two 

 tents during twenty minutes would require in one case a doubling and 

 in the other case a fourfold increase. The ratio between these two 

 rates of increase is that one is the square root of the other. 



The formula would be 



dose d^ X <^\/l 



in which d = diameter and 1 = leakage factor. 



Our unit diameter is 10 ft. and the leakage factor that satisfies the 

 condition of the average dosage practice is four as shown by the fol- 

 lowing calculation: 



