# Ballistic Re-entry and the effects of lift

1. Aug 4, 2011

### bobster4562

Hey guys,

I understand that 'ballistic re-entry' is usually used in the context of drag only with 0 lift, but if I were to look into the effects of lift how would I go about resolving the forces acting on a body at each instant to determine the instantaneous body acceleration ?

My initial thoughts are - resolve instantaneous velocity V into its X and Y components (say Y is North-South pole and X is perpendiculer to it).

Lift being perp. to V means Lx and Vx would be in the same direction, but Vy and Ly would be in the opposide direction. I could then do

Acc x = gx + Lx - Vx [-Vx is basically subtracting the drag force)
Acc y = gy - Ly - Vy

gx, gy being the gravitational acceleration.

Am I on the right lines here ?

Cheers !
Bob

2. Aug 4, 2011

Not especially. You have units all wrong. V and L don't have units of acceleration, for starters.

Finding the lift itself is also pretty complicated. You will need to know the pressure distribution over the body and thus the velocity distribution over the body.

3. Aug 4, 2011

### bobster4562

My apologies, I was a bit careless when writing the equations down -

By L I mean the acceleration due to lift force where

L = 1/2 x rho x v^2 /(2 x Beta_L)

and V I'm implying the acceleration due to drag where

D = 1/2 x rho x v^2/(2 x Beta_D) [and D acts opposite to V]

Beta_D = m/(A x Cd) where

m = mass, A = aerodynamic area and Cd = drag coefficient

However I am not sure if I could apply an equivalent expression for Beta_L - initially Im thinking I could use a similar expression but I'm not sure. I do not want to over-complicate the problem since I appreciate that the final result (i.e. the effect of lift for ballistic re-entry) would be very small. What do you guys think ?

Thanks !

4. Aug 4, 2011

The problem is that no matter how you slice it, the problem is complicated.

Anyway, you do have a few conceptual problems. First, lift and drag are not really separate forces. They are the components of a single aerodynamic force vector which represent the force exerted by the air against the direction of motion and the force exerted by the air normal to the line of motion. In other words, your force balance equation will look like this:

$$\Sigma F_x = a_x = mg_x + F_d$$
$$\Sigma F_y = a_y = mg_y + F_l$$

The sign of each will depend on how you define your coordinate system and what direction your object is moving in that coordinate system. Generally it is best to work with a right-handed coordinate system, so let $x$ be along the instantaneous line of motion and let $y$ be normal to that in a manner such that $|\hat{x} \times \hat{y}| = 1$.

Finding the value of $F_d$ and $F_l$ is generally nontrivial and how you go about it depends on how accurate you want to be and what tools you have at your disposal. In general, they will not be simply velocity dependent, but instead dependent on many things such as altitude, orientation, shape, velocity, etc.

5. Aug 4, 2011

### bobster4562

Thanks, I appreciate your point. At the moment I am considering y to be earths N-S axis and x to be perpendicular to it, so all my forces would be resolved in this coordinate system.

Assuming my body is descending in the +X, -Y direction (ie to the bottom right corner of your screen)

would you agree if I said that in my X direction the associated components of drag and lift would oppose each other while in my Y direction the components would add up ?

This is what Im really interested at, just to make sure Ive got my vector additions correct !

Many thanks !
Bob

6. Aug 4, 2011

I would agree based on your coordinate system, however, using a body-centered coordinate system would be much simpler and is generally how real fluid calculations are performed.

7. Aug 11, 2011

### bobster4562

Cheers, would also agree that this approach ie -

L = 1/2 x rho x v^2 /(2 x Beta_L)

D = 1/2 x rho x v^2/(2 x Beta_D) [and D acts opposite to V]

Beta_D = m/(A x Cd) where

m = mass, A = aerodynamic area and Cd = drag coefficient

Beta_D = m/(A x Cl) Cl = lift coefficient

would be a reasonable way of modelling the drag and lift forces acting on the body ?

8. Aug 11, 2011

### rcgldr

Those equations don't work well for supersonic speeds, unless Beta_L and Beta_D are defined as functions of velocity, Beta_L(v), Beta_D(v), in which case, you've really changed the relationship between drag and lift versus density and speed. For external ballistics, which involves speeds much slower than re-entry speeds, tables are often required in order to get accurate drag numbers:

http://en.wikipedia.org/wiki/External_ballistics

You're coordinate system, is cylindrical, where x is really r, the distance from the axis of the cylinder (north south axis). Spherical coordinates might be better for an explanation, where the "slope" of the reentry path could be described as dr/ds (change in radius versus change in distance). The actual math might involve a mix of cartesian coodinates (x, y, z) and spherical coordiantes (latitude, longitude, altitude).

Last edited: Aug 11, 2011
9. Aug 11, 2011

### bobster4562

For ballistic re-entry case, Im assuming Cl does not change that much (we are talking a few minutes or so of re-entry flight time I guess). I appreciate that there is incredible amounts of drag and hence Cd must be a function of velocity or derived from a table or something.

10. Aug 11, 2011

### rcgldr

The questionable article about the Columbia breakup linked to below includes some basic equations (apparently ones taught in early aerodynamics classes), where lift and drag coefficients above mach 2.5 and at relatively low density of the upper atmosphere are functions of the angle of attack. This conflicts with the chart from the wiki external ballistics article which shows a decreasing trend in coefficient of drag at mach 2.5, but that is at near sea level densities of the air. I'm not sure if the stated coefficients for lift and drag are accurate, but the form of the equations seems correct.

I only posted this link as a source for those basic equations to give you an idea of the math involved. The rest of the article appears to be about some conspiracy or cover up which I didn't notice, until pointed out by boneh3ad below, so I would recommend ignoring the rest of the article.

http://www.columbiassacrifice.com/\$A_reentry.htm

The article mentions in a foot note It is unknown why the increase in velocity between the end of the deorbit burn and EI (Entry Interface) is so great. If it is not an error it may by a byproduct of the entry process. The fact that the deorbit burn puts the shuttle into an elliptical path that has decreased altitude by 530,000+ feet should explain the large increase in velocity from deorbit burn completion to entry interface. This makes me wonder about the accuracy of the article overall, but again, the form of the equations does give an idea of the math involved. You'd need to find more accurate values (tables) for the coefficents of lift and drag versus speed and altitude.

As an alternate link, here is a nasa article:

http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19690003968_1969003968.pdf

I also found a mill spec link, but the actual article link seems to be broken:

Last edited by a moderator: Apr 26, 2017
11. Aug 11, 2011

That is not a good assumption. Both the drag and the lift will change dramatically based on many variables such as velocity, density, spacecraft attitude, spacecraft shape, and other factors.

That columbiasacrifice.com site looks to me like a crackpot conspiracy theorist site claiming that either a nuclear explosion or HAARP set off an EMP-like burst that downed the shuttle. Their aerodynamic equations are those taught to students taking their very first aerodynamics course and are incredibly simplified. The actual situation is much more complicated.

The NASA paper seems to be more of a flight controls paper, as does the second link. They seem to be interested in using the lift to the advantage of the spacecraft, but they aren't actually doing the aerodynamics.

Last edited by a moderator: Apr 26, 2017
12. Aug 11, 2011

### rcgldr

The term ballistic reentry normally means only drag is involved. If lift is used, the term normally used is lifting reentry.

I updated my previous post to note that the overall article was questionable, and that the equations were simplifications, but would give an idea of what was involved.

That is the reason I posted the link to that questionable article, since I couldn't find those simplfied equations elsewhere, although I seem to recall seeing similar equations in the past.

This gets back to finding a reasonable source for simplified reentry equations, which I could not find other than that questionable article. These wiki articles don't include any math, and the second link is just a stub.

http://en.wikipedia.org/wiki/Atmospheric_entry

http://en.wikipedia.org/wiki/Ballistic_reentry

Another NASA article with a lot of math:

http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19640016000_1964016000.pdf

Last edited: Aug 11, 2011
13. Aug 12, 2011

I don't think you can find a simplified version. It is a necessarily complex problem. That is one of many reasons why only a handful of countries are spacefaring.

The simplest place to start would probably be Newtonian impact theory. It is a very good estimate at entry speeds.

14. Aug 12, 2011

### bobster4562

The last NASA article is very interesting - they do not use a constant lift but a constant-ish L/D ratio. Thanks very much !

15. Aug 12, 2011