*Copyright 2019 Robert Clark*

*In this tweet I speculated on the possibility that the SpaceX's BFR upper stage, or BFS for Starship, when given wings might need no thermal protection at all:*

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I speculated on this possibility here: https://mobile.twitter.com/rgregoryclark/status/1094210607740567553?s=21 … The impetus was this report that a space plane at very low wing loading (weight per wing area)would not need thermal protection: Wings in space. by James C. McLane III Monday, July 11, 2011 http://www.thespacereview.com/article/1880/1

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https://twitter.com/RGregoryClark/status/1117752403946889216

https://twitter.com/RGregoryClark/status/1117752403946889216

The idea behind it was an article that suggested with sufficiently low wing loading, weight per wing area, an orbital stage might need no thermal protection at all on reentry:

by James C. McLane III

Monday, July 11, 2011

*Wing loading (the vehicle’s weight divided by its wing surface area) is a prime parameter affecting flight. The antique aluminum Douglas DC-3 airliner had a big wing with a low loading of about 25 psf (pounds per square foot of wing surface). At the other end of the spectrum, the Space Shuttle orbiter has a high wing loading of about 120 psf. This loading, combined with an inefficient delta-shaped wing, makes the orbiter glide like a brick. A little Cessna 152 private plane features a wing loading of about 11 psf and modern gliders operate down around 7 psf. A space plane with huge lifting surfaces and a very low wing loading might not require any external thermal insulation at all. Building a space plane with a wing loading of, say, 10 psf should not be an impossible proposition. Perhaps some day it will be done.*

The stage intended as the Mars colony ship with the passenger quarters and provisions for 100 colonists on a 6 month trip to Mars, the BFS, gave a pounds per square foot at above twice the 10 psf level:

According to Wiki it has a dry mass of 85 tons, with a diameter of 9 m and length of 55 m. The lateral surface area of a cylinder is the circumference times the height (by imagining it sliced down the middle and flattened out.) The circumference is Pi times the diameter. However, the surface presented to the area stream when reentering broadside would be half the full area. So this area would be 0.5*Pi*9*55 = 777.5 sq.m. = 8,365.1 sq.ft.

The mass of 85,000 kg is 187,000 pounds, giving a pounds per square foot of 187,000/8,365.1 = 22.4 psf.

Elon Musk may or may not have been raising the possibility (it's not always clear when Elon is being serious) of giving the BFS wings to get it below the 10 psf value with this tweet:

However, the tanker version, with no passenger quarters and just an empty fairing, might only weigh ca. 50 tons, based on a comparison of its size to the earlier, larger Interplanetary Transport System. This is a weight in pounds of 110,000 lbs. Using the same size for the tanker, the area presented to the air stream would still be 8,365.1 sq.ft, and the pounds per square foot would be 110,000/8,365.1 = 13.1 psf.

This might be workable to only need a small additional wing area to get it beneath the 10 psf. number.

However, it is interesting that some existing upper stages are below or close to the 10 psf number. For instance the Delta IV Medium upper stage:

Converting the weight to pounds and the lenghts to feet, the psf is:

2.2*2,850/(.5*Pi*4*12.2*3.28^2) = 7.6 psf, less than the 10 psf requirement.

The larger upper stage used on the Delta IV Heavy would also satisfy the 10 psf requirement:

The psf is:

2.2*3,490/(.5*Pi*5*13.7*3.28^2) = 6.6 psf.

The Atlas V's Centaur upper stage also satisfies the 10 psf requirement:

The psf is then:

2.2*2,243/(0.5*Pi*3.05*12.68*3.28^2) = 7.55 psf.

Note that the RL10 engine used on the Delta IV Medium, Delta IV Heavy, and Atlas V is both restartable and highly throttleable to be used for a vertical, propulsive landing:

In fact the degree of throttleability is such that it can do a

*hovering*landing.
This is important for propellant saving on landing. SpaceX does a hoverslam approach to vertical landing because of the low degree of throttleability of the Merlin engines. This wastes a significant amount of fuel because the thrust has to be so much larger than the nearly empty weight of the stage.

In fact for powered landing of a stage the required fuel reserve mass is commonly taken to be only in the range of 10% of the

*landed, i.e., dry mass of the stage*. See for example the discussion here:
It's because even without wings a stage reentering broadside would burn off almost all the orbital velocity by atmospheric drag alone. The result is the stage would only be at terminal velocity at 100 m/s as it approached landing which would have to be cancelled by out by the engines. With deep throttling the required thrust, and so propellant burned, would be much reduced so as to only cancel the nearly dry weight of the stage.

The Falcon 9 also nearly reaches the 10 psf number:

The psf is:

2.2*4,000/(0.5*Pi*3.66*12.6*3.28^2) = 11.3 psf.

Then so small wings or fins could be used to bring it under the 10 psf range. However, a bigger problem is the vacuum Merlin engine does not have high throttleability and being vacuum optimized, it can not operate at sea level for the landing. In upcoming posts I'll discuss methods to solve these problems.

Actually, small movable fins might be needed in any case, for the other cases as well. This is because upper stages tumble and break-up when they reenter. They would have to be actively controlled to maintain the same attitude presenting their lateral side to the airstream to burn off the orbital velocity and to reduce stresses that would cause them to break up.

The planned movable fins on the BFS might be used:

Another possibility is the variable flaps at the rear of the ESA's IXV vehicle:

Bob Clark

2.2*1,240/(0.4*

**UPDATE, 7/1/2019:****In the blog post "ESA's Callisto reusability testbed as an *operational* TSTO and SSTO", I suggested the Ariane 4 H10 upper stage be used for the Callisto reusability testbed, with its engine swapped out for the Vinci engine. It turns out this stage as well satisfies the 10 psf requirement for reentry without thermal protection. It had a 1,240 kg dry mass, 2.60 m diameter, and 11.05 m length. So the psf is:**

2.2*1,240/(0.4*

*π**2.60*11.05*3.28^{2}) = 5.6 psf .
I looked at something similar, posted over at my "exrocketman" site, as "Pivot Wing Spaceplane Concept Feasibility", dated 3 April, 2019. My craft sized out as 60 lb of entry weight divided by body cross-section with wings folded dead broadside (NOT nose-on to the flow).

ReplyDeleteThe main difference is in the vehicle inert weight. In order to carry a folding wing mechanism, folding landing gear, and enough structural "beef" to survive entry airloads dead broadside, I used inert structure some 20% of gross weight at ignition. It came out at 60 psf, not your target 10 psf, or anywhere close.

I ran my ballistic entry spreadsheet, and figured peak gees as reasonable, and added a re-radiation estimator that gave me 2350 F equilibrium peak stagnation surface temperature at the peak heating point, with a black surface emissivity of 0.80.

2350 F is way too high even for stainless steel, and for the turbine and afterburner super alloys. Aluminum is "right out". I was looking to stay under 2350 F in order to use low-density ceramic even at the stagnation point.

My sizing came out this way precisely because of a complex tradeoff amongst entry ballistic coefficient, how much folding wing I could package onto a body of a given size with the pivot concept, and feasible landing speed at sea level (under 200 knots).

My inert weight fraction of 20% may seem high to you. It is probably only slightly conservative, and only because the wings are folded completely out of the hypersonic flow during entry. Spacex's low inert weight fraction for its Starship looks suspiciously unrealistic to me, even for entry angles of attack limited to under 40 degrees off dead nose-on.

I had done a hinge-type folding straight wing feasibility study earlier, also posted on "exrocketman". The numbers were similar, except I could get to a much lower landing speed. Entry was roughly the same. That one is "A Unique Folding-Wing Spaceplane Concept", dated March 2, 2013.

GW

Thanks for that. Can you run a hypothetical scenario with a psf of 10? Also, what reference works do you use for the equations that go into your spreadsheet?

ReplyDeleteBTW, there are some grades of stainless steel with melting point above 2,700 F. But you would still have to worry about its strength weaking at 2,350 F.

Bob Clark

SS 309/310 resists surface oxidation to 1900 F, but is no stronger than 304 at 1200-1400 F, namely under 1 ksi tensile. That stuff is Mil Handbook 5 stuff, or manufacturer's datasheet stuff. You use only the annealed strength data if you go hot, with any of the 300 series.

ReplyDeleteMy spreadsheet for entry uses the same equations H. Julian Allen used in 1955+/- for warhead entry. In my day, they still taught this stuff in graduate school aero/heat transfer courses. In the 1950's and 60's it was classified, but was declassified in the 70's.

I just put it to work again, and corrected some errors from the reference I used (I no longer use the closed-form solutions for peak heating or accumulated heat or for peak gees). That reference was the Justus & Braun EDL report, and they goofed up converting Allen's work to metric units.

Yeah, I can run a 10 psf loading in my entry spreadsheet. What drag coefficient should go in the denominator? It would likely be a number between 1 and 2 for almost any conceivable shape, with total vehicle planform area as the reference area. I presume you are talking about dead-broadside (90 degree AOA) entry.

Note that I have serious doubts your wings could possibly stay attached for a dead broadside entry, not at weights corresponding to a 10 psf "wing loading". Peak gees will tell you something about that.

If you want a copy of my entry spreadsheet, I could email it to you. The key variables are speed at entry, angle below local horizontal at entry, and entry interface altitude, plus the ballistic coefficient and "nose" radius of the entering object. The scale height and atmosphere factors for Earth are pretty well fixed, although with other values, you can model any planet.

GW

Titanium-Aluminium alloys might be the best option here.

DeleteThanks for that. I’ll send you my email address for the spreadsheet.

ReplyDeleteAbout the drag coefficient I get the feeling the author of the “Wings in Space” article was literally talking about the (weight)/(wing area) ratio, i.e., no drag coefficient included.

Normally the quantity discussed is the ballistic coefficient. This is (weight)/(drag coefficient*wing area). I don’t know what the hypersonic drag coefficient is for a cylinder entering broadside. For a sphere it’s about 1 for supersonic speeds, so perhaps also at hypersonic speeds. As a first estimate we can take it also as about 1 for the cylinder. But try to calculate the ratio without the drag coefficient since I think that’s what the author meant.

About the movable fins staying attached at hypersonic speeds, Elon has discussed that difficulty as well as moving them at those speeds. I gather SpaceX believes it is solvable.

BTW, here’s another reference for low ballistic coefficient reentry:

Applications of Ultra-Low Ballistic Coefficient Entry Vehicles to Existing and Future Space Missions.

https://spacecraft.ssl.umd.edu/publications/2010/SpaceOps2010ParaShieldx.pdf

But this has an annoying “typo”. The report uses metric units. On the first page it gives an equation for the ballistic coefficient that has mass in kg’s in the numerator, not weight which would be in newtons. But then he uses pascals as the units for the ballistic coefficient which means he IS using the weight in newtons in the numerator.

He does give a discussion about the decision whether to use mass in kg’s or weight in newtons in the numerator of the ballistic coefficient in a footnote on the first page. But he could have eliminated the confusion just by using the weight in newtons in that first equation.

Bob Clark

Ballistic coefficient is formally defined in terms of mass per unit (blockage area x an appropriately-referenced drag coefficient). Wing loading is actually defined in terms of weight per unit lifting area. Depending on the object, the areas themselves are just not the same.

ReplyDeleteIf you interchange weight and mass, you must use inconsistent units, and you are thus restricting yourself to Earthly scenarios. Terribly inconvenient, that.

I'll run you an entry spreadsheet at low ballistic coefficient. We'll see what the peak stagnation point surface temperature looks like, and the peak gees.

-- GW

Suicide burn is more efficient than hover landing.

ReplyDeleteIts also accurately named...

Actually it isn’t. The reason is because you have to burn so much more fuel when your T /W > 1, than when your T/W = 1.

DeleteDo a calculation where you want to cancel out a 100 m/s terminal velocity with T/W = 1. You’ll find the required propellant reserve is only in the range of 10% of the *dry mass*. So only about 400 kg for the upper stage, and 2,500 for the first stage.

For the propellant reserve for the Falcon 9 landing where it is landing off-shore, i.e., no return to launch site, which is the relevant scenario, see the discussion here:

https://www.reddit.com/r/SpaceXLounge/comments/7ifn09/how_much_fuel_does_the_f9_ft_really_need_to_land/

The estimate is ca. 10% of the full propellant load, or ca. 40,000 kg for the first stage(!)

Note also that the Blue Origin’s New Shepard is able to hover and does do a hovering landing. Granted this might be because of wanting an extra level of safety, but if you think about it, why burn more fuel, i. e., thrust higher, than you have to?

Bob Clark

I disagree with this.

DeleteIf you are burning propellant to produce a TWR of just 1, then you will not slow down as gravity cancels out your thrust and therefore velocity is unaffected.

Let's try for a TWR of 1.3 for stage mass of 2500kg. With an Isp of 310s, you need just 80.8 kg of that mass to be rocket fuel to cancel out 100m/s. It takes you 7.8 seconds... right?

No! You are also falling at 9.8m/s^2 during that time. This means only 0.3 of your 1.3 TWR is actually slowing you down. In other words, the 7.8 second burn experiences a 'gravity loss' of 77m/s! This increases your fuel consumption to 141kg.

Let's increase TWR to 2.6. Fuel consumption per second doubled... but the burn time is halved. No change in total fuel required to accelerate by 100m/s. What about gravity losses? They are halved too, because you spend less time fighting gravity with your engine.

You’re right that gravity losses are reduced by thrusting at T/W > 1 initially. But at some point, it is advantageous to reduce the T/W to T/W = 1.

DeleteThe Falcon 9 booster landing off-shore and the New Shepard rocket are both suborbital. But the Falcon 9 fuel reserve requirements are widely out of whack with the 10% of the dry weight estimate commonly given for the fuel requirements of a vertically returning stage, while the New Shepard’s are in concert with it.

Find some landing videos of New Shepard. Then estimate the fuel usage for the landing burn from the equation (fuel-burn rate) = (thrust)/(Isp*g)

SpaceX could greatly reduce the payload loss on reusability by giving the F9 booster capability for T/W = 1.

Bob Clark

Matter Beam, that link in the post to the discussion of the propellant requirements for vertical landing was incorrect. I changed it now to the correct link:

Deletehttps://yarchive.net/space/launchers/horizontal_vs_vertical_landing.html

Bob Clark

Matter Beam, I was assuming the difference in payload loss from the SpaceX hoverslam landing and the usual 10% of dry mass loss estimate for a hovering landing was because the Falcon 9 booster didn’t hover.

DeleteBut that’s not it. The key reason is that even without return-to-launch-site they apply a reentry burn on the booster to limit heating on reentry. That surprised me as I assumed that the booster being only suborbital did not need to limit reentry heating. For instance New Shepard does not need a reentry burn.

I also found that on the boosters final approach, where the final landing burn is applied, the speed is not just the terminal velocity of 100 m/s. It’s closer to 300 m/s. That surprised me as well. This means it needs a larger burn than would be the case for just 100 m/s.

This seems to imply that the booster does not do a broadside reentry. This brings us back to the issue of using a broadside reentry to limit reentry heating. The questions is for the slower speed of a suborbital reentry can the broadside approach limit heating so that the aluminum-lithium tanks can withstand it?

I’ll discuss this in a follow up post.

GW, you might want to chime in on this as well.

Bob Clark

Thanks for the additional information. 300m/s sounds high bit it does seem plausible for the terminal velocity of a giant structure pointed end-on into the airflow.

DeleteAlso, I think the limiting factors for a broadside re-entry are:

-Structural stresses. Wind buffeting at supersonic speeds might fold it in half!

-Aerodynamic stability. The CoG is far below the CoL, making it stable end-on but very hard to keep horizontal.

-Flip danger. Swinging from horizontal to vertical is a risk where the rocket can roll/yaw violently and in the wrong direction.

I found a gmail address for you that I hope still works. I used it to send you the entry spreadsheet file. I ran a low ballistic coefficient comparison study for you. It's in the spreadsheet.

ReplyDelete-- GW

Thanks. Got it.

DeleteBob Clark