96 M. A. F. Sundoll on Absolute 



If we put these values in the equation given, we have 



1 gramme (Paris) =c a x 0*001 kilogramme x 9*808 - — - 2 -, 



sec. 



i /-o • \ n nAnQA o metre x kilogramme 



1 gramme (Pans) = 0*009808 — f x c; 



and 



1 sec. 2 x gramme (Paris) 

 C ' 2 ~ (K)09808 metre x kilogramme 



The rule given above is also of service in the exchange of 

 units, as is shown in the following examples : — 



Example 5. — Let the velocity of light be taken as funda- 

 mental unit instead of the unit of time. We know that 



Velocity of ligkt = 40,000 ^g^aphical miles . 

 J ° second 



Then we have the velocity 



1 metre _ 1 geogr. mile _ 1 . _ .. , 



^ond ~ 7420 second 7420 x 40^00 Ve g f 



and the acceleration 



9*808 ^ = 9.8O8 (met f x l 



(sec.) 2 (sec.) 2 metre 



9*80 8 (velocity of light) 2 



metre 



From this last number we may obtain the ordinary value as 

 follows : — 



9*808 vel. of li ght 2 _ 9*808 x 40,000 2 geo gr. mile'- ' 



7420 2 x 40,000 2 metre ~~ 7420 2 x 40,000 2 sec. 2 x metre 

 9*808 x 40,000 2 x 7420 2 _ metre 



7420 2 x 40,000 2 ~ sec. 2 ' 



Example 6. — Let the acceleration due to gravity (Paris) 



= 9*808 5- be taken as the fundamental unit instead of the 



sec. 



unit of time. We obtain then, for example, 



„ , __ metre X kilogr. (Paris) --met* . i 



1 horse-power =75 ^V =75 x metre 2 



1 second sec. 



x kilogr.(Paris) = 75( \ J x met.'* x kilogr. (Paris). 



= —7=== acceleration due to gravity (Paris) 2 x metre* 

 V 9*808 . „ ' ._ . 



x kilogr. (Pans). 





