Mach 10, Top gun

Thread Starter

drjohsmith

Joined Dec 13, 2021
1,563
Just been to see Top Gun II with my son
To be honest , was not expecting much , my "memories" of the original in a BIG screen in central London, was fantastic
but must admit, the follow on is quiet good fun, instantly forgettable, but fun

Anyway
without spoiling, There is a Mach 10 plane in the film, that does a nice turn at over Mach 10 with our hero piloting.
So question is ,
how big a turn would that be ?

Im guessing lots of variables,
after all its meant to be an impossible plane,
but assuming a human, they can take say 3 g for a fair while ? or is that to high / low
and a Mach 10 at some great height , is 100 000 feet reasonable,

just how wide would a 180 degree turn be,
miles ? 10 of miles ? to me that still feels tight, but who knows,

Any one got the equations "to hand" to take a stab ,
Thank you
 

SamR

Joined Mar 19, 2019
5,472
There is an old story about an SR71 pilot contacting air traffic control requesting permission to break 80,000'. Air traffic controller came back to him with a snide "just how do you think you are going to get up TO 80,000'"? There was a short pause then "Uhhh... I'm DESCENDING"! So 100,000' could be possible, but Mach 10? The top speed for Ram Jets is ~Mach 6 and they really poop out over Mach 5. Even manned space launches only get to 3G, sooo... But then they are done by NASA, not MGM...
 

Thread Starter

drjohsmith

Joined Dec 13, 2021
1,563
Thanks guys
Yep the Mach 10 plane in topgun us fictional.
Anyone know of an equation for turn radius, speed and g force ?
 

Papabravo

Joined Feb 24, 2006
22,058
The speed is not the important consideration. It is the bank angle which correlates to the g-forces and the rate of turn. Doing a shallow turn at Mach 10 is probably not a problem, but a turn at progressively higher bank angles will result in ever increasing g- forces. Assuming the pilot is wearing a g-suit, 9 g's is probably the limit assuming the airframe does not fail.
 

BobTPH

Joined Jun 5, 2013
11,466
The radial acceleration formula is:

a = v^2 / r

r = v^2 / a

set your acceleration to 3 g or 3 x 32ft / sec^2
and speed to Mach 10 or 10 x 1125 ft / sec

and you get:

r = 1e8/ 96 or about 1 M feet or 197 miles! Not exactly turning on a dime.
 

ApacheKid

Joined Jan 12, 2015
1,762
If you never watched The First Of The Few then give that movie a try. It's about Mitchell the skilled aircraft designer from the days when Britain dominated international speed competitions, who went on to design the Spitfire prior to the outbreak of WW2. Yes its a bit dated but a great and true story.

Tragically the actor who played Mitchell in the movie was killed shortly after the movie was finished, in a passenger plane that was shot down accidentally by the Germans (it did really seem to have been unintentional too).
 
Last edited:

Thread Starter

drjohsmith

Joined Dec 13, 2021
1,563
The radial acceleration formula is:

a = v^2 / r

r = v^2 / a

set your acceleration to 3 g or 3 x 32ft / sec^2
and speed to Mach 10 or 10 x 1125 ft / sec

and you get:

r = 1e8/ 96 or about 1 M feet or 197 miles! Not exactly turning on a dime.
Thank you Bob p.
That was perfect
Exactly what I wanted .
197 miles of 3g
That would be fun.... Not
Again thank you
 

BobTPH

Joined Jun 5, 2013
11,466
Thank you Bob p.
That was perfect
Exactly what I wanted .
197 miles of 3g
That would be fun.... Not
Again thank you
I think I experienced that on the “roundup” carnival ride once. At my age (about 60 at the time) that was not fun.

It is a a rotating circle that starts horizontal and goes full vertical. So if you don’t fall off at the top, it is more than 2g at the bottom.
 

nsaspook

Joined Aug 27, 2009
16,258
I think I experienced that on the “roundup” carnival ride once. At my age (about 60 at the time) that was not fun.

It is a a rotating circle that starts horizontal and goes full vertical. So if you don’t fall off at the top, it is more than 2g at the bottom.
 
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