Apparent discrepancies in atmospheric reentry in SFS (or, Is it hot in here or is it just me?)

§PaCëØddît¥

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#1
After experiencing the new reentry effects introduced into the game (thank you Stef!) a couple of things struck me as a little.... off kilter.
In the tangible world the effects of the top 10% of a planets atmosphere is negligible, it's basically still a near vacuum, but in SFS the effects are almost immediate and spectacular.
Don't get me wrong, there's lots of things going on to cause these effects. A kilogram of matter re-entering the Earth's atmosphere at 1,600 m/s in SFS has 1.28 * 10⁶ joules of energy, so take your average capsule, parachute, heat shield package of 4.9 tons we get a whopping 6.272 * 10⁹ joules of energy. but like the ronco guy says, wait! there's more! This is just the kinetic energy, we've still got the little matter af the crafts mechanical energy to deal with as well. what's that you ask? well our craft is still sitting on top of a 30 km tall cliff. We calculate this little number and we add an additional 1.47 * 10⁸ kg/m or 1.442 * 10⁹ joules to the mix for a total of 7.713 * 10⁹ joules of energy to deal with.
Well hell yeah! that would explain the light show and with that much energy being thrown around we must be burning through velocity like mad! Right? Turns out that isn't quite the case. For the purposes of this discussion I quick saved a 4.9 ton craft entering Earth's atmosphere at an angle of 20° for repeatability as I gathered data.
Turns out that the light show, despite being quite visually arresting doesn't have the same effect on our crafts velocity. I've got graphs galore but just showing the numbers at key points of our reentry shows some of the issues.
Firstly, at 27 km our temperature is ≈ 200° C and velocity has crept up to 2118 m/s in keeping with the real life model (kudos Stef!). Let's look at the data as we progress , 30 seconds in our temperature is 850° C progressing mostly linearly with some flattening to 2130° C at 90 seconds into reentry. At our peak energy consumption we're only consuming 7.89 * 10⁷ joules a second, certainly enough to power a light show but barely putting a damper on our velocity.
Alright, now we know what we're dealing with what can we do with it? Since we neglected to carry the equivalent fuel required to get us here and meany Stef won't let us deploy our parachute in a vacuum or at supersonic speeds, we're going to need something to slow us down.
There are 2 basic atmospheric reentry approaches, ballistic and lifting. A lifting approach entails entering an atmosphere at a shallow enough angle of incidence the craft exits the atmosphere rather than continuing in the direction of the retarding force.
We're going to examine the ballistic approach.
Ever since cannoneers began to analyze the trajectories of cannonballs we've quantified the forces acting on an object traveling through the atmosphere.

!!! CAUTION MATH AHEAD !!!

The 3 basic formulas I applied are:
equation of motion

a = ½ P * v² * Cᵈ * A
m

where a is the angle of flight, P is atmospheric density, v is velocity, Cᵈ is coefficient of drag, I used the Cᵈ of the Mercury capsule 1.6, A is area and m is mass

ballistic coefficient. A measure of a body's ability to overcome air resistance in flight.

B ᶜ = m
Cᵈ * A

where m = mass, Cᵈ is coefficient of drag and A is area

and finally, Fᵈ the drag force on our vehicle.

Fᵈ = ½ P * V² * Cᵈ * A

So if we look at the data now the numbers begin to align. The key factor being P, atmospheric pressure.

Running the numbers again, we see, at the beginning of our entry, due to minimal atmospheric pressure the effects of drag are negligible;

Fᵈ = ½ .000001 * 2,114 m/s² * 1.6 * 3.14 = 22.45 joules. Once we get into the thick of things, so to speak, we find that joules, and velocity, bleed off rapidly

Fᵈ = ½ .8 * 1,644 m/s * 1.6 * 3.14 = 5.431 * 10⁶ joules and our velocity ( and the force required to slow us) is dropping rapidly.

In conclusion, if you've made it this far, all is well with the laws of physics in the SFS world in regards to atmospheric reentry. (apologies Stef but I had to check)
I believe my misperception was caused by a time dilation effect the visuals created. I simply perceived time to be moving much faster than it actually was.

My apologies for any errors in notation, an unfortunate limitation of my keyboard. Any errors in math are my own as well.

references include but are not limited to:

Wikipedia, the free encyclopedia › wiki
Atmospheric entry - Wikipedia

Conquering the fast and the furious: Physics of atmospheric re-entry – theGIST

https://www.faa.gov › mediaPDF
Returning from Space: Re-entry - FAA
 

Altaïr

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#2
Cool, some maths :p

That looks mostly correct, though SFS is a game, there may be approximations compared to real life.

Regarding the atmospheric density, I don't know if the values used are some random ones, but it's possible to calculate it for a given height in SFS. For this, you need those parameters:
Screenshot_20220624-191440_QuickEdit+.jpg

"height" is the atmospheric height, "density" is the air density at ground level (expressed in tons/m³), and "curve" a parameter that determines how fast density decreases with altitude. The formula is:
ρ(h) = density × exp(-curve × h / height)

Where ρ(h) is density at height h.

For Earth, the density is halved every 2200 meters approximately.

By the way, I do a lot that mistake too, but the drag formula uses density, not pressure. This is a current abuse of language, but density and pressure are two different things. Density is the mass per unit of volume (in kg/m³), pressure is the force exerted by the atmosphere per unit of surface, in pascals, or N/m².

Regarding the heat generated during reentry, I believe it's around a few MW/m² at the peak in real life. Your value of 7.89×10⁷ J/s (so 78.9 MW) seems a bit high, but you would have to divide by the "surface". Also, most of the heat is dissipated into the atmosphere during reentry, the few MW/m² I am talking about are the part that goes into the heat shield, so this seems consistant.

This is hard to compare with real life though. The reentry speed in real life is 7800 m/s, not 1660, and I guess you see how that changes V²...
 

§PaCëØddît¥

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#3
I'm having a real issue with atmospheric density. What I'm getting with atmosphere height of 30 km and a curve of 10. Density is .5 at 2000 meters .2 at 3000 meters and thins to nothing rapidly after that. Big problem. No atmosphere = no drag = no heat
I'm reworking the numbers but my first hypothesis is, unfortunately, holding true. There's simply nothing in the top 10 kilometers of atmosphere to generate the temperatures shown.☹️
 

§PaCëØddît¥

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Regarding the heat generated during reentry, I believe it's around a few MW/m² at the peak in real life. Your value of 7.89×10⁷ J/s (so 78.9 MW) seems a bit high, but you would have to divide by the "surface". Also, most of the heat is dissipated into the atmosphere during reentry, the few MW/m² I am talking about are the part that goes into the heat shield, so this seems consistant.

I was trying to avoid Newton/meters even though there's a "direct" correlation. Just the whole force versus energy thing. Energy over time equals work(force) The total energy converted from kinetic to heat isn't the sum total being (here I don't believe dissipated is correct) converted.
There are other real life factors that aren't being considered here. The bow shock wave creating its own insulating layer. Also the shield is being treated as a heat sink type ( reusable) versus ablative.
These issues are relatively minor. Absolutely,atmospheric density is issue. Unless someone can explain it. There's just no way the math works. It looks like someone plugged in the temperature requirements and filled in the values to get them. Looks great on the surface but just doesn't work. Though in the real world you truly don't want to attempt an an areobraking maneuver anywhere except the very fringes of atmosphere. An aerobraking maneuver used by NASA to insert a vehicle in orbit around Mars took 6 months. Around earth this maneuver would begin around 250 kilometers .
 

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§PaCëØddît¥

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#5
Regarding the atmospheric density, I don't know if the values used are some random ones, but it's possible to calculate it for a given height in SFS. For this, you need those parameters:
View attachment 85626

"height" is the atmospheric height, "density" is the air density at ground level (expressed in tons/m³), and "curve" a parameter that determines how fast density decreases with altitude. The formula is:
ρ(h) = density × exp(-curve × h / height)

These are the settings for the default solar system in the game?

If so then the math does not work. Very disappointed.

I am glad to see new parameters added. Just at a quick glance it would appear that an atmosphere structured to much more closely resemble Earth's, Albeit on a 1/20th scale is possible. Maybe I'm missing something in the physics that wouldn't allow this? Old parameters Screenshot_20220624-212830.png

At this point I am continuing calculations to see if all kinetic and mechanical energy is accounted for. I just can't get the numbers right for the crafts interaction with the atmosphere's edge. More study of spacecraft re entering earths atmosphere in real life might help.
 

Altaïr

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#6
I finally spotted a mistake here:
well our craft is still sitting on top of a 30 km tall cliff. We calculate this little number and we add an additional 1.47 * 10⁸ kg/m or 1.442 * 10⁹ joules to the mix for a total of 7.713 * 10⁹ joules of energy to deal with.
The formula you used for potential energy is the common form:
Ep = m×g×h

But that works supposing that g is constant, which is fine if h is small compared to the Earth radius. But here it's not the case, at 30 km the gravity already dropped by more than 15%!

Fortunately the exact formula is still simple:
Ep = - μ × m / (R + h)

μ is the gravitational parameter (G×M in the usual astrodynamics formulas). For Earth ingame, you have:
μ = 972.405 × 10⁹ m³/s².
R is the Earth radius (315000 m for Earth in SFS).

And yes, it's negative, but because potential energy is a relative term. What counts is the difference of potential energy between 2 points. So here the potential energy is actually given by:
ΔEp = Ep(h = 30000m) - Ep(h = 0m)

You get ΔEp = 1.315×10⁹ J instead of 1.442×10⁹ J.

This doesn't drastically change things though, since kinetic energy is the dominant factor.

I'm having a real issue with atmospheric density. What I'm getting with atmosphere height of 30 km and a curve of 10. Density is .5 at 2000 meters .2 at 3000 meters and thins to nothing rapidly after that. Big problem. No atmosphere = no drag = no heat
I'm reworking the numbers but my first hypothesis is, unfortunately, holding true. There's simply nothing in the top 10 kilometers of atmosphere to generate the temperatures shown.☹️
You're correct, the density decreases exponentially, and it becomes quickly very low.

But you shouldn't conclude that there's nothing to generate drag.

First thing, maybe check your calculations because I don't know how you get those numbers. Density at ground level is 5 kg/m³ (0.005 is in tons/m³), so at 2000 m, it is:
ρ(2000 m) = 5 × exp(-10 × 2000 / 30000) = 2.57 kg/m³.

At 10 km, ρ = 0.178 kg/m³
At 30 km, ρ = 2.3 × 10^(-4) kg/m³

At a high altitude density is very low, but in the drag formula, it is multiplied by V². So at 1660 m/s, the dynamic pressure is:
P_dyn = 1/2 × ρ × v² = 312.8 N/m²

You see it's not negligible. You only have to multiply by the surface and the Cd coefficient to get drag.

For the heat generated it's worse. The power generated by drag is P = Fd × v. If we assume that the power is fully converted to heat, we have:
P = 1/2 × ρ × v³ × Cd × A

Power is proportional to v cubed!

It doesn't look like but you can quickly get some huge values. In particular, the formula P = Fd × v implies that at a high speed, even a low drag will be enough to generate a significant amount of heat.

A tragic example of this is the Columbia shuttle, that was destroyed when it was still at 60 km above ground level. It looks like we are still far from space at 60 km, but the density is already 10000 times lower than at ground level there. It's only significant because of the speed at which reentry occurs, a human exposed to that atmosphere would die very quickly the same as if he was in vacuum.
 
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§PaCëØddît¥

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#7
My apologies for the delay getting back.

First a couple of corrections.

Initial re-entry velocity is 2118 m/s not the 1600 m/s used in the original calculations. Recalculating total energy of our craft - kinetic energy = to 4,900kg × 2118 m/s = 10. 378 × 10⁶ kg/m = 10.378 × 10⁶ × 9.804 = 1.018 × 10⁸ Joules + Ep = μ × m / (R + h) @Altair = Ep = 1.35 × 10⁹ Joules for a total energy of 11.528 × 10⁹ Joules.( Although I got total energy a complete factor lower at 2.165 × 10⁸ joules)

This is the energy we must account for for the math to work.

Another error was my reading of the Fd formula. I read it as ½ p × V² × A. Correctly it's Fd = p × V²/2 × A. Density times velocity squared divided 2 × A.

Additionally, the behavior of the atmosphere interacting with an object travelling at supersonic speeds should be included. This will include air flow as well as air column compression in front of the craft as these are significant factors. Speed of sound @ 20°C = 343m/s @ -60°C = 292m/s. Craft velocity at atmospheric entry = mach 7.25. The calculations show supersonic flight characteristics don't come into play

Formulas:

Density

D = Cd ×( p × V²) × A
2

Isentropic ( Ice-n-tropic) Compression

T2T1 = (P2P1) y - 1y

T1 = initial temperature, T2 = temperature at the end of the compression process, P1 = initial pressure, P2 = compressed pressure, y = 75 for diatomic gas.

R = Hs @ Pc - Hs @ Vc

R = specifc heat at a constant pressure (cp) minus the specific heat at a constant volume (cv) = the gas constant.

A brief word on an Isentropic process. There is no transfer of heat because it's adiabatic. (process or condition in which heat does not enter or leave the system concerned.)


https://physics.stackexchange.com › ...
Compression vs. Friction - what happens when space objects enter Earth's ...

I apologize if this is confusing. I realize something that is perfectly clear to me from this side may be confusing as hell from your side. I'm attempting to lay out all the formulas and definitions I've used to reach the conclusions I have so that any who follow might have a foundation to work from.

There's 2 forces we're going to be addressing. Drag and friction. First, drag, this is perhaps the most eloquent definition I've run across

Drag - a force (kg/m² , Newtons) acting opposite to the relative motion of any object with respect to it's surrounding solution. This can exist between 2 fluid layers or between a fluid and a solid layer. Unlike other resistive forces, such as friction, drag force is velocity dependent. It is proportional to velocity at low speeds and squared to velocity at high speeds.

Reynolds number is a function of the craft, a low number produces laminar flow, a high number, turbulence, with an accompanying consumption of energy.

A re-entry vehicle is subject to both friction and drag

In the real world several kilometers of atmosphere are compressed into a thin layer of gas in front of the craft in less than a second. It's this compression that produces the majority of heat.



Data collected- Height(km), Velocity (m/s), Temperature °C), Deceleration (m/s), Atmosphere Density (kg/m³), Drag Force (N), Friction(N). All data collected at 15 second intervals unless otherwise noted. Length of data collection period - 270 seconds resulting in 19 collection points. I attempted to maintain a straight line in every collection run.

I went ahead and calculated energy converted per second for the entire 270 second re-entry.

Appears I need to work on my XML.

Total energy accounted for 5.886 × 10⁷. Off by a complete factor, 2 factors if we go by the higher total sum energy. Let me get my data organized/presentable.
 

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Altaïr

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#8
My apologies for the delay getting back
No worries, we all have a life.

I also remember I tried to understand the physics behind reentry, but I broke my teeth on it (french expression). It's really complicated, it requires a specialist at some point.

The reentry physics ingame are really simple in comparison. I believe the formula for heat generation is something like:
ΔT = 1/2 × ρ × V^(2.5)

The formula for heat generation I gave above was 1/2 × ρ × V³, so I suppose Stef used this but he wanted to give less importance to the speed factor... There's also a cooling process, so that the heat shield gradually loses the accumulated heat.

Regarding the thermodynamics, I'm not an expert there. Too bad, we had an expert at this on the forum once ago... :rolleyes:
 

§PaCëØddît¥

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Greetings from Sweden
My conclusion to atmospheric re-entry in SFS.

While not completely realistic, the physics are internally consistent.

Certain aspects are ignored. Superheating, atmospheric compression, etc. This does not invalidate the process. Indeed, kudos to Stef and all involved devs for maintaining uniformity within the game.

First and foremost SFS is a game, and as such, fun, entertainment as it were, is a primary goal. For those that want to turn on every cheat available and galavant among the stars no one should look askance. We've all used a cheat or two on occasion.
Just please bear with those of us who dabble in the science side such as it is. View attachment 88432
 

Mooncrasher

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That's a lot of data.:eek: I'm sure there's a lot of interesting analysis that can be done with it.

How did you collect the data? Calculated or empirical? Particularly the drag data.
 

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That's a lot of data.:eek: I'm sure there's a lot of interesting analysis that can be done with it.

How did you collect the data? Calculated or empirical? Particularly the drag data.

It's not so much analysis as usefulness. The formula for drag is Fd = p × V²/2 × A. p = atmospheric density (not pressure) times velocity squared divided by 2 × A (area of craft presented to atmosphere)
IMG_20220809_140341805.jpg


This is just one of many, many runs. If you go back through this post I'd hope more would be made clear
 

§PaCëØddît¥

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It's not so much analysis as usefulness. The formula for drag is Fd = p × V²/2 × A. p = atmospheric density (not pressure) times velocity squared divided by 2 × A (area of craft presented to atmosphere) View attachment 89049

This is just one of many, many runs. If you go back through this post I'd hope more would be made clear
I lost a ton of data moving around