Calculation of the Rate of Climb

An important performance parameter is the rate at which an aircraft can increase altitude and the time needed to achieve a particular altitude. Estimates for these quantities are determined readily by using tools that we already assembled. We con­sider Fig. 9.6, which show an airplane in climbing configuration. The angle 9 between the velocity vector and the horizontal is the climb angle, which obviously has an important role in the calculation. The force balance in steady-climbing attitude indi­cates that we must have:

jT = D + W sin 9 (9.17)

l = W cos 9, (918)

where lift and drag still are given by the usual formulas. Notice that the thrust force now must carry part of the weight as well as balance the drag.

What is required is the vertical speed, or rate of climb, (RC). This is clearly given as the component of flight speed in the vertical direction:

RC = Esin 9. (9.19)

Equation 9.17 can be used to determine RC by solving for and then multiplying through by the flight speed. The result is:

RC = TV – DV. (9.20)

W

Recalling that force times speed is the power (i. e., rate of doing work), a useful phys­ical interpretation of Eq. 9.20 as well as a practical method for calculating the RC, is obtained. The products of thrust and drag with velocity are the power available and the power required, respectively. Their difference often is called excess power—that
is, the power available to climb or to execute other maneuvers. If the excess power is zero, then an airplane is flying at maximum speed in unaccelerated, level attitude. Therefore, we see that the RC can be written as follows:

RC = excess^power. (9.21)

A pilot controls RC by changing excess power by means of the engine throttle.

Подпись: dPR dV Calculation of the Rate of Climb Calculation of the Rate of Climb Подпись: (9.22)

The maximum available RC can be estimated easily if either numerical data or a mathematical expression for the power available is known. In Fig. 9.5, a simple model for PA is illustrated. It assumes that the power available is not affected significantly by flight speed or altitude. In fact, there are usually rather important sensitivities to the speed and air density. In actual practice, we would have the detailed infor­mation available. However, we can use the simple model to illustrate the procedure. Referring to Fig. 9.5, we see that the maximum excess power corresponds to the minimum point in the power-required curve because the power available is assumed to be insensitive to speed. Therefore, it is necessary to estimate this low point and to determine the corresponding flight speed. Comparison of Figs. 9.4 and 9.5 shows that the minima in the thrust required and power-required curves do not occur at the same flight speed. The necessary information is found by setting the derivative of the power required with respect to speed equal to zero to locate the extremum. The value of speed corresponding to minimum power required is found by solving:

Calculation of the Rate of Climb Calculation of the Rate of Climb Calculation of the Rate of Climb Подпись: (9.23)

for this special value of V. The result is:

which shows that the speed to fly for maximum RC is about 24 percent less than the speed to fly for minimum power required (i. e., maximum L/D). Thus, for the Bf – 109G, the best climbing speed at sea level is approximately 163 ft/sec (111 mph). The speed for best RC increases with altitude approximately as the inverse square root of the ratio of density to the sea-level density, as indicated in Eq. 9.23 and verified in Fig. 9.5. Because the power available decreases similarly with altitude, it clear that the RC diminishes with altitude.

Inserting the best climb speed into Eq. 9.15 yields all of the necessary informa­tion for determining RC and climb angle, as summarized in Table 9.5. The results shown agree closely with published data for this aircraft.

Table 9.5. Climb performance of the Bf-109G

Altitude

(ft)

Speed, V (mph)

Excess Power* (horsepower)

Climb Rate, RC (ft/min)

Climb Angle, 0 (degrees)

sea level

111

863.8

4,255

23.5

22,000

154

791.8

3,900

16.1

* Assumes that constant power available = PA = 0.88 • 1,200 = 1,056 horsepower.

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