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So I noticed on X-Plane 11, the SR-71 reaches Mach 3 at 80000 feet, but the equivalent airspeed is only around 200 knots. What explains this difference between the mach number readout and the EAS readout?

Edit: the reason I ask is because other questions such as How did SR-71 spy, flying at 80,000 ft and 3500 km/h? indicate the SR-71 flew at 1910+ knots at 80000 feet, but I'm not sure if this is with a conversion of some sort or not, i.e., is this simply a simulator problem or an accurate representation of airspeed and mach?

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    $\begingroup$ related: aviation.stackexchange.com/q/3693/1467 $\endgroup$ – Federico Nov 12 '17 at 19:23
  • $\begingroup$ You can use this calculator. Under standard atmosphere, at 80,000 ft the speed of sound is 579 kt (661 at sea level), but density is only 0.043 kg/m3 (1.23 at sea level). That explains why the indicated airspeed is very low. $\endgroup$ – mins Nov 12 '17 at 21:18
  • $\begingroup$ I've worked out an answer to this. I get indicated airspeed = 401 m/s = 780 kts. I cannot post the answer because while I was composing it, using this equation, the question was closed. The cited questions for duplicate do not list a direct answer to this question. Requesting that this question be re-opened please, so I can post the answer. $\endgroup$ – Koyovis Nov 13 '17 at 4:12
  • $\begingroup$ Thx for re-opening. Equation above is for incompressible flow only, got a value in order of magnitude of X-plane. $\endgroup$ – Koyovis Nov 17 '17 at 14:41
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    $\begingroup$ Maybe you should call for a ground speed check... you'd get an accurate answer, and it would quiet down any fighter jocks nearby. $\endgroup$ – tj1000 Nov 17 '17 at 16:11
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Indicated airspeed is derived from the measured total pressure and the static pressure, according to:

$$ V_i = \sqrt {\frac{2 \cdot (p_t - p_s)}{\rho_{SL}}} \tag{1}$$

At 80,000 feet, static pressure is 2761 Pa. Dynamic pressure as measured by a pitot tube, is measured after a supersonic shock wave according to the Rayleigh Pitot tube formula:

$$ \frac{p_t}{p_s} = \left[\frac{{(\gamma + 1)}^2 \cdot M^2}{4\cdot\gamma \cdot M^2 - 2(\gamma - 1)} \right]^{\gamma / (\gamma - 1)} \cdot \frac{1 - \gamma + 2\cdot \gamma \cdot M^2}{\gamma + 1} \tag{2}$$

with $\gamma = 1.4$ and M = 3, substituted in (2), gives $\frac{p_t}{p_s} $ = 12.06, so $$p_t = 12.06 * 2761 = 33\,290 ~\text{Pa}$$. Substitute this into (1) together with $\rho$ = 1.225 at sea level, and we get

$$ V_i = \sqrt{\frac{2 \cdot (33,290 - 2761)}{1.225}} = 158 ~\text{m/s} = 307 ~\text{kts}$$

In the order of magnitude of what the X-plane indicated airspeed indicates, a lot closer than the TAS: speed of sound @ 80,000 ft = 298 m/s, Mach 3 = 894 m/s = 1,738 kts

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    $\begingroup$ Very cool – so if I follow, the short answer is yes, the readouts on X-Plane are more or less accurate for Mach 3 at 80,000 ft, being much lower than the TAS. Incidentally, since the Blackbird seems to measure EAS instead of IAS, I should have earlier referenced Equivalent airspeed, which makes for a much easier calculation, essentially multiplying the TAS by about 5 to get the EAS at 80,000 ft. So the numbers on the simulator all seem to check out. $\endgroup$ – wander Nov 17 '17 at 20:10
  • $\begingroup$ Is 307 kts for EAS or IAS? $\endgroup$ – wander Nov 19 '17 at 1:52
  • $\begingroup$ The 307 would be for IAS. $\endgroup$ – Koyovis Nov 19 '17 at 14:50
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I already knew the speed of sound decreases with increasing altitude – I was actually trying to figure out why the Blackbird Wikipedia specs list a speed of 1910+ knots at 80,000 ft when the EAS only read 200-300 knots at said altitude. The conversion between EAS and TAS explains the difference.

It turns out I made a simple problem out to be harder than it was. Since the Mach numbers of the Blackbird are non-trivial, EAS is used instead of IAS, and per Equivalent airspeed, using

EAS = TAS * sqrt(p/p0)

with p being the actual air density at 80,000 ft of .043 kg/m^3 and p0 being the standard sea level density of 1.223 kg/m^3, making the EAS roughly a fifth of the TAS. So a TAS of Mach 3, 1,738 knots, would have an EAS of 326 knots. I assume, then, the simulator is more or less accurate.

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  • $\begingroup$ Equivalent Air Speed at Mach 3 is not that simple to convert. The equation you're using is valid for subsonic flows that may be considered incompressible and for which we can take dynamic pressure $ p_d = \frac{1}{2} \rho V^2$ $\endgroup$ – Koyovis Nov 18 '17 at 1:21
  • $\begingroup$ Are any of the equations listed on the Wikipedia page usable then? $\endgroup$ – wander Nov 19 '17 at 1:52
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    $\begingroup$ It's a bit of a hard one. There are lots of references to the EAS for incompressible, subsonic flow, not so many for high supersonic flow. When I look it up in this Flight Testing manual eq. 4.28 I get a much higher value. $\endgroup$ – Koyovis Nov 19 '17 at 15:18
  • $\begingroup$ Referring to a much higher value for the EAS for Mach 3 at 80,000 ft? $\endgroup$ – wander Nov 21 '17 at 4:41
  • $\begingroup$ Yes indeed. In trying to get to the bottom of it I got stuck in trying to figure something out on supersonic dynamic pressure $\endgroup$ – Koyovis Nov 21 '17 at 5:28

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