Why the angle of attack is 40 degrees for shuttle started to enentry

Why the angle of attack is 40 degrees for shuttle started to reentry

I'm a newbie here. Because I am doing an assignment on the Columbia Shuttle Disaster. I need some data for supporting my statement.
The shuttle must then tilt upward to present its flat bottom, covered with heat-resistant ceramic tiles, to the upper atmosphere, with the nose angled upward at 40 degrees, the so-called "angle of attack." This angle is critical to the shuttle's re-entry. If it is greater than 40 degrees, the craft could flip over backward from the atmospheric thrust. Less, and it would risk entering the atmosphere too fast, and overheating and possibly melting its aluminum shell.
Thus, I wonder how to calculate initial angle for shuttle reentry is 40 degrees? How about the resistance and gravity acting on the spacecraft? And what's the direction of velocity for it enter the atmosphere?

Thanks a lot!

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Answers and Replies

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Whoahboy, that's not an easy question...

You'd need to actually work on the program to know the formulas for calculating the reentry path.

How much physics have you had?

I can come up with some generalities for you, I just need to know how technical to make it...

To make the 40 degrees more clear is okay.
I think I can understand yours. May you give me some information about these so far as you known? Even if you give me the basic thing which just like the velocity direction for reentry. I will try to analyze it by myself

Zero degrees gives almost no braking - thus they can hit Earth with a speed of a metheor - not much fun.

90 degrees gives too much braking - acceleration quickly reaches -20g - also not much fun (if you survive it).

How to calculate it? Setting acceptable value of acceleration (say, -3g).

Because atmosphere density is increasing with descend, angle shall be decreasing too with descend.

Of course, actual optimum angle versus h function will differ from calculated due to many small things which are impossible to predict (wether jets, actual density of atmosphere due to Solar activity, etc) thus this angle shall be constantly adjusted by wing tails during re-entry.

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When you are moving through an atmosphere at supersonic speeds, shockwaves form in front of your air(or space)craft. If you are flying an airplane, you want to have very little drag. To accomplish this, you keep the edges as sharp as possible (fighter aircraft wings can pose a hazard to the ground-crew because they are so sharp).

If you are coming into the atmosphere in a spacecraft, you want drag. The air is supposed to slow you down. That is why the noses on all spacecraft are blunt. It's also why the space shuttle is pitched. By pulling back on the stick and coming in at an angle, you effictively increase your profile, increasing the drag.

The reason why you don't pull back farther is to keep the shuttle stable. (I'm winging this... I could be wrong here)
Coming into a flow straight on causes what are called 'normal shock waves'. That means shock waves that are at 90 degree angles to the flow. Angled shock waves are called 'oblique'. Oblique shock waves form when a wave hits a non-perpendicular disturbance.

In the shuttle in re-entry, you will have a normal shock in front of the nose and oblique shocks down the bottom. The less oblique you are, the lower the workable pressure you have at the surfaces. They are really low for normal shocks... the flow actually goes subsonic also (this is why supersonic jet intakes have the points out front). If the shuttle would come in at a higher pitch, then the oblique shocks along the bottom edge would get more normal, reducing the pressure on the bottom (and wings). The lowered pressure would cause the nose-down torque due to the wings to lower, and the nose-up torque due to the nose would overpower it. This would cause the shuttle to flip and disintegrate because un-heat-shielded parts would then be facing the 2000 degree air.

I wrote an article on www.physicspost.com about hypersonic (greater than Mach 5) flight a few months ago. It has a few pictures of normal and oblique shocks and a very basic overview of what happens when you pass through them.

http://www.physicspost.com/articles.php?articleId=93" [Broken]

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Thank you very much!
The information is very valuable!

I think the point is that one wants the re-entry path as long as possible in order to be able to slow down the shuttle gradually without creating too much over-heating and g-forces. The path has to be therefore very shallow; in fact, one could say that the space shuttle flies almost horizontally as it takes 8000 km to drop by just 130 km. What one needs for this is an aerodynamic lift which is about equal to the drag; for a plane surface, this is the case at an angle of attack of 45 deg (the drag causes the shuttle to lose horizontal speed; this would lead to a rapid loss of height and increase of vertical drop speed if it is not offset by an equal amount of lift). Of course, the exact shape of the shuttle would modify this value somewhat, which could explain the value of 40 deg.

P.S.: I have derived the 45 deg value in my post Bernoulli's Law and Aerodynamic Lift (page 2, second post)

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I found it!

dV/dx = V' = g sin(&alpha;) - F/M
http://home.att.net/~numericana/answer/physics.htm#reentry [Broken]

For proper re-entry the angle of reentry a cannot be too small, or else the spacecraft could "bounce" off the atmosphere and be back into outer space after losing just a little bit of energy. If no action is taken, an orbiting spacecraft could thus keep bouncing back until enough energy is lost and/or its reentry angle is sufficiently large --possibly dangerously so, since a large a means fast reentry and a lot of heat!

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Don't forget that shuttle is a plane - it can't miss it landing site even a few miles, because there is only 2 of them around - in FL and CA, (it can't land just in any airport because it needs quite long runway). So it can't bounce back and forth from atmosphere and finally drop somewhere into Pacific or Siberia like regular landing modules do. It can't even slightly miss narrow window of landing parameters for its destination.

Therefore the angle (and the point of re-entry) shall be accurately set (and gradually decreased according to density of air, as shuttle descends) and constantly adjusted according to projected by computer landing point.

Also, due to high deceleration (3-4g) during re-entry the crew is not very capable of doing much if things go wrong. Just physically (imagine operating in a hurry complex machinery while riding roller-coaster).

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Originally posted by Sorsby
dV/dx = V' = g sin(&alpha;) - F/M
http://home.att.net/~numericana/answer/physics.htm#reentry [Broken]
The angle 'a' in this formula is not the 'angle of attack' of the shuttle but the angle of the flight path relative to the earth's surface. The two are only identical if the shuttle is orientated horizontally, i.e. the nose pointing towards the horizon. However, the nose is actually pointing 40 deg above the horizon, i.e this is the angle of attack, but the shuttle is flying almost horizontally to the ground, i.e. 'a' in the above formula is close to zero.

But you could probably use this formula to calculate the required deviation of the angle of attack from 45 deg for a given slowdown, as 45 deg would mean the speed stays constant (see my post above).

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Your derivation is incorrect. Drag and lift are vectors directed at completely different directions, so they can not cancel each other.

Also you completely neglected aerodynamics which actually provides most of lift.

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Originally posted by Alexander
Your derivation is incorrect. Drag and lift are vectors directed at completely different directions, so they can not cancel each other.
Also you completely neglected aerodynamics which actually provides most of lift.

I did not say they cancel other, but 'offset' each other.
Drag causes a loss of speed (which you want here), but a loss of speed would cause a loss of height which again would increase your speed (because of energy conservation). So with a corresponding amount of simultaneous lift, the drag will slow down the shuttle without leading to a height loss at the same rate.
This is an argument of aerodynamics here as the angle of attack determines both the drag and lift, but you can also consider for instance the case of a satellite subject to friction in the upper atmosphere: for a spherical satellite there is no lift at all (only a drag force opposing the velocity) and the satellite will both lose height and gain speed at the same time (half of the gravitational potential energy will be converted into kinetic energy). You could only prevent the speeding up of the satellite by providing some lift to it.

Originally posted by Alexander
Don't forget that shuttle is a plane - it can't miss it landing site even a few miles, because there is only 2 of them around - in FL and CA, (it can't land just in any airport because it needs quite long runway). So it can't bounce back and forth from atmosphere and finally drop somewhere into Pacific or Siberia like regular landing modules do. It can't even slightly miss narrow window of landing parameters for its destination.

You're wrong. Re-entry process only starts in the ionosphere just after de-orbit burn; and ends in the Mesosphere. Then shuttle will be controllable.

You can see:

How Space Shuttles Work
http://science.howstuffworks.com/space-shuttle.htm/printable

The danger starts at one hour before landing
http://www.spacedaily.com/2003/030202084210.y9712m23.html [Broken]

Earth's Atmosphere
http://www.physics.usyd.edu.au/~cairns/teaching/lecture16/node2.html

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I know it's rehashing a very old thread, but we had a very brief intro to re-entry maneuvers in one of my classes today.

When a shuttle, or any other spacecraft, begins its re-entry maneuver, the vehicle is in an extremely rarified atmosphere. The mean free path of an air molecule (the average distance a molecule will travel before hitting another molecule) is much larger than the length of the spacecraft. For this reason, the 'collisions' can -and are- modelled as discrete impacts - "Newtonian flow".

By drawing a free-body diagram of a flat plate and considering the change in momentum of the plate is equal to the change in momentum of the molecule, the lift and drag work out to be:

L= 2 * &rho; * V2 * A * sin 2 &alpha; * cos &alpha;
D= 2 * &rho; * V2 * A * sin 3 &alpha;

giving lift and drag coefficients of:

Cl = L/( .5 * &rho; * V2 * A) = 4 * sin2 &alpha; * cos &alpha;
Cd = D/( .5 * &rho; * V2 * A) = 4 * sin3 &alpha;

If you graph the lift and drag vs. angle of attack, you find that the maximum lift is obtained at an angle of attack of ~55 degrees - the angle of the shuttle's re-entery.

They do this to increase maneuverability and to expand the range of landing options.

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