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This equation is of the form V4 +a V +b = 0, where
                                                          k'r7p P'(kW)
                                  a - -                                   (2.112)
      pSCDO
        /7= C (g)2      (2.113)
    The equation  V4 + aV + b-0 has no closed-form analytical solution. We
have to obtain either a numerical or graphical solution to determine Y~x  For this
purpose,let us plot R/C against velocity as shown in Fig. 2.15. This plotis called
the hodograph of climbing flight. A hot71ograph is a graph in which the variation
of one velocity component is plotted against the other.
   The maximum climb angle occurs at that velocity when the excess thrust per
unrt weight is maximum. This corresponds to the point where a line drawn from
the origin is tangent to the hodograph as shown in Fig. 2.15.
     The time to climb from a given initial altitude h, to final altitude hf  is given by
      ,v dh
t-   -                 (2.114)
  .-i Pr
R/C
AIRCRAFT PERFORMANCE
O
Fig.2.15   Hodograph ofcLimbing flight.
97
which is equal to the area under the curve obtained by plotting inverse specifrc
excess power against altitude as shown in Fig. 2.16. A special case ofinterest in
climbing flight is the minimum time to climb. We will discuss this in more detail
when we address the same issue for ajet aircraft as the approach will be essentially
sinular for the propeller aircraft
      Ana/ytica/ so/utions for c/imbing f/ight of jet aircraft.    Ifwe assume that the
  thrust developed by ajet aircraft is independent of flight velocity, then it is possible
 to obtain some analytical solutions climbing fiight as follows.
       Assuming that the climb angle y is small, Eq. (2.61) reduces to L - W or n - 1,
 and, using Eq. (2.26), Eq. (2.62) takes the following form:
          zW  W
                             E   - 2V/t. (.2+' ) - W siny =0                 (2.115)
or
                                siny = 2~. [2z - (.2+' )]                     (2.116)
For y to be a maximum,
                                            d sin y
      d =-2E~(2u-2)
                                    = 0                                    (2.117)
or
                                                           u -. 1                                            (2.118)
　
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