Mr. S. Sharpe on the Figure of the Cells of the Honeycomb. 21 



Fig. 3. the plain end of the above. Join the alternate an- 

 gles ACE to the centre g, and to 

 each other. Then if the centre g Fig. 3. 



be raised and ACE remain fixed, 

 the other angles (solid angles in 

 the prism) BD and F will be 

 lowered, and the solid contents of 

 the cell will remain the same. 



Again, by comparing fig. 1 

 and 2. we see that though the py- 

 ramid needs more materials than 

 the plain termination, yet the sides 

 of the former prism need less than 

 those of the latter by the six little 

 triangles Amb, 



Now let AB(=mb)be a 



A m (the altitude of the pyramid) ... x 

 AH (half of AC) d 



Ab= V a* + x* 



Amb = — - 



2 



Hb = s/ cC l + x*—d\ and the quantity which 

 we wish to be a minimum is 



3(AgCb)-6(Amb) = 6d \/ a* + x t — d i ]- 3 a x whose 



fluxion is SdQ* + x 2 — d 2 ^xx — 3 ax — 

 2dx(a t +x i -d 2 Y h = a 



2dx = a(a?+x*' 

 4.d*x 2 z=;a 4 +a?x 2 



■a 2 d? 



ai—cfld 1 

 Ad* -a* 



but d* = a*-£- = 



3(1* 

 x = 



therefore x*= 



3a« 



4 



3 a* -a* 



and 



•T 



Ab = a v-g-j 



and Ab : AH : : radius : cos. HA b = 





'A 



COS 



35° 15' 51"-86, and the angle g Ab = 70° 31' 44" nearly. 



AgC = 109 28 16. 



VII. Che- 



