# Why does X-Plane 11 show the SR-71 at 200 knots but Mach 3 at 80,000 ft?

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?

• – Federico
Nov 12, 2017 at 19:23
• 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.
– mins
Nov 12, 2017 at 21:18
• 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. Nov 13, 2017 at 4:12
• Thx for re-opening. Equation above is for incompressible flow only, got a value in order of magnitude of X-plane. Nov 17, 2017 at 14:41
• Maybe you should call for a ground speed check... you'd get an accurate answer, and it would quiet down any fighter jocks nearby. Nov 17, 2017 at 16:11

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

• 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. Nov 17, 2017 at 20:10
• Is 307 kts for EAS or IAS? Nov 19, 2017 at 1:52
• The 307 would be for IAS. Nov 19, 2017 at 14:50

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.

• 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 $p_d = \frac{1}{2} \rho V^2$ Nov 18, 2017 at 1:21
• Are any of the equations listed on the Wikipedia page usable then? Nov 19, 2017 at 1:52
• 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. Nov 19, 2017 at 15:18
• Referring to a much higher value for the EAS for Mach 3 at 80,000 ft? Nov 21, 2017 at 4:41
• Yes indeed. In trying to get to the bottom of it I got stuck in trying to figure something out on supersonic dynamic pressure Nov 21, 2017 at 5:28

All this calculated stuff is impressive. I just figure the air density at 80.000 is less dense and it doesn't have enough pressure to force the airspeed indicate up. Being it takes air pressure in the pet tube to cause the airspeed indicator gauge to go up in numbers. Even though it's only shows 200 knots, it's speed over the ground is much faster. So I'm thinking it could be close to stalling if it go's any higher unless it can push it's self faster. If I remember corect the 71 rotation speed is 180 knots. This is a interesting subject. I know the space shuttle shows no airspeed until it inters the atmosphere. I wounder what the numbers are on it.

• You are describing indicated air speed. you can improve your answer by putting this term in it. Apr 7, 2020 at 6:32