TM 1-320  19-20  AIR CORPS  where  Op  is a constant which can be taken from the compilation  below:  b.  Solving the problem given in paragraph 17g (1) by the Burgess  formula gives- 20. Speed developed by g iven horsepower.-a. By transpos› ing the horsepower for mulas the following formulas are obtained for  the speed developed by a given horsepower :  b. Problem and. solution.-(1) Problem.-An  airship of 195,000- cubic-foot capacity has a power installation of two motors developing  150 horsepower each, or a total  of  300 horsepower.  The atmospheric  density is standard.  What speed should be obtained at full powed  (2)  S olution.- Using  Burgess’ formula.  34  3  /H.  P.  X  01)  v= -y  p(vol)21a  3  I  300 X 20000  - v  o.oo237 x (1 95ooo) 2 ’ 3  =90.8 feet per second= 6L9 miles per hour.  $^{2}$.sn  /H.  P. X550XEX F  (H. P.X550 XEXF\ 0 ’ 36  v=-y  CvXPX  (vol)2ta  =  CvXP  X (vol)2/3 - ) from paragraph  17}.  $^{3 }$/H. P.  X 550 X E  =  -y  0 ,  X  P  (  vol )2’3 from paragraph 17  g.  3fH.P. X 01)f  h  9  = V  P  (  vol) 213  rom paragrap  1  a.  P  (  vol )213  va  H. P.=  Cp  - (0.00237) (195,000) 11 ’ 3  (88) 8  20,000  = 273 horsepower.  R igid  airships  1,000,000 to 2,000,000 cubic feeL _______________________ Op=30,000  2,000,000 to 3,000,000 cubic feeL- ---------------------- Cp=32,000  3,000,000 to 4,000,000  cubic feeL _______________________  Cp=33,000  4,000,000 to 6,000,()()() cubic feet_ _______________________ Cp= 34,000  6,000,000 to 10,000,000  cubic feeL____________  -------- Cp=35,000  NO"nrigid  airships  .  50,000 to 200,000 cubic feet--------------- ------------- Op=20,000  200,000 to 300,000 cubic  feet_ ________ .. .  -•---------- -- Op=21,000  300,000 to 400,0J:>  cuui<: feeL--------•-•----------------- Op= 22,000 