What is the minimum turning radius of an SR-71 at Mach 3.2 and an altitude of 80,000 feet?

I have heard that if an SR-71 were to cross the Pacific coast over San Francisco and pull a hard turn to the right, it would return over Seattle.

I'm interested in real-world performance, taking into account things like inlet unstart (failure to capture the engine inlet shockwave in the intake) from too sharp an angle of attack.


4 Answers 4


The Turn Radius of the SR-71 would depend on its speed. The faster it went the wider its turn radius was.

The SR-71 had a minimum turning radius at altitude of about 80 nautical miles (NM) . It was not an airframe limitation but a matter of wing area. At 80,000ft, the air is too thin and the wings too small to allow for much lift to turn with.

At a turn radius of 80 NM, the SR-71 would cover about 145 miles, taking about 4 minutes in the process of making a 180 degree turn.

Details are provided in the SR-71 flight manual handbook, which is now declassified: Fig. 1-9, SR-71 handbook

  • $\begingroup$ What is the turning radius at mach 3.2? $\endgroup$
    – user2168
    Commented Aug 6, 2014 at 19:20
  • $\begingroup$ Check Figure 1-9 books.google.co.in/… $\endgroup$
    – DSarkar
    Commented Aug 6, 2014 at 20:39
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    $\begingroup$ @D_S Why? Is that where the answer is? If so, it should be in your answer here, too. $\endgroup$
    – user2168
    Commented Aug 6, 2014 at 22:41
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    $\begingroup$ i looked at the figure. It does not list the minimum turning radius at 80,000 feet at mach 3.2. Including your interpretation of the relevant portion of the figure here is not a violation of copyright. $\endgroup$
    – user2168
    Commented Aug 7, 2014 at 8:06
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    $\begingroup$ The top answer states: "At a turn radius of 80 NM, the SR-71 would cover about 145 miles, taking about 4 minutes in the process of making a 180 degree turn." This should raise the reader's eyebrow. If the turn radius is 80nm, then after completing a 180 degree turn, the plane would be laterally offset by 160nm (184 statute miles) from its previous track. Its actual ground track would cover 251 nm (almost 290 statute miles) in that 180 degree turn. $\endgroup$ Commented May 3 at 16:12

Please use the equations of this answer. The numbers might be different, but the physics are the same.

EDIT: Thanks to D_S for providing the link to the manual.

When flown with the maximum allowable load factor of 1.5 g at 80.000 ft (48° bank), the turn radius at Mach 3.2 (equivalent to v = 953.3 m/s in 80.000 ft) will be 83.5 km. To be more precise, you will need to add the effects of earth rotation, but for now I leave this away. As you can see, the turn will still need 163 km or 103.7 miles, but not the distance from San Francisco to Seattle which is more than 6 times bigger.

To turn this around: A circle at Mach 3.2 which has a diameter of 1092 km requires a bank angle of 9.6°. That can hardly be called a turn.

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    $\begingroup$ The equations are nice from a theoretical standpoint, but you need to pick your assumptions when using them. The SR-71 is so far outside of "normal" for airplanes that I don't know what assumptions are reasonable. $\endgroup$
    – Mark
    Commented Aug 6, 2014 at 9:31
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    $\begingroup$ @Mark: I refined my assumptions, but the SR-71 was not so unusual that it would violate the laws of motion. Those equations are valid for all airplanes. $\endgroup$ Commented Aug 6, 2014 at 9:44
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    $\begingroup$ The equations are valid. But the maximum possible load factor is an argument we need to know. The load factor not only needs to be allowed, but also achievable at cruise altitude and speed. $\endgroup$
    – Jan Hudec
    Commented Aug 6, 2014 at 12:37
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    $\begingroup$ @Jan Hudec: At Mach 3.2 the possible lift is out of the question - increase angle of attack, and the lift will be there. The main problem is the drag increase which could be more than the engines can compensate (and the efficiency of the intake at the different angle of attack). 1.5 g should not be an issue, however, if the airplane was flying level with some margin before. All it needs is half empty fuel tanks, and there is plenty of margin already. $\endgroup$ Commented Aug 6, 2014 at 13:58
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    $\begingroup$ @Articuno: More than that. The diameter grows with the square of the speed (please look up the formulas in the first link in my answer!). His 80 NM at Mach 3.0 translate to 91 NM at Mach 3.2 or 104.7 miles. This leaves just a rounding error between both results. My answer is both physically and legally correct, and actually answers Mark's question. $\endgroup$ Commented Aug 7, 2014 at 16:19

Considering the current flight status of the SR-71 (retired) it's maximum speed will depend on the tug moving it and the turn radius is probably around 30-40 meters depending on how far the nosewheel pivots.

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    $\begingroup$ I would like to see a tug that can pull an SR-71 at Mach 3.2 and 80,000 feet. $\endgroup$
    – fooot
    Commented Aug 6, 2014 at 16:44

I worked one in about 1989 that was flying eastbound toward Rapid City, and started his turn back toward Edwards. He rolled out over GLL VORTAC, so about 200 nautical miles south of where he started. He was at roughly 2400 knots (He slowed down for the turn). Altitude was above FL600, but we did not get information on actual altitudes.


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