Stoke Space

[tater]

 

[magnemoe]

On 2/23/2023 at 9:47 PM, tater said:

I meant the whole stage 2, so both tanks, sorry, wasn't clear.

Yep, I wasn't even looking at that end, lol.

Not seeing the heat shield from this angle, though... maybe they will do hot fires, then install it? (easier to work on engines minus everything being buried under that metal).

And playing the Dodd interview back skipping around made it pretty clear it was as I remembered—they were standing next to the hopper doing the interview.

You can see the payload adapter base plate to the right and to the left its looks like the stuff going under the heat shield. Still can not come over how small upper stage is and they are using hydrogen. 
 

[JoeSchmuckatelli]

On 2/25/2023 at 11:33 AM, tater said:

 

Wen hop?

(That was cool - I hope they succeed!)

[tater]

4 hours ago, JoeSchmuckatelli said:

Wen hop?

(That was cool - I hope they succeed!)

I'm very much a fanboi for these guys. Of all the newspace launch providers, these are the first with a really novel idea.

(excluding SpaceX, the all-in-one Starship concept is novel)

[tater]

They got a pad:

 

[SunlitZelkova]

For such an innovative company to get Glenn’s pad is the most awesome thing ever.

[Superluminaut]

I think this is the only company currently pushing beyond spacex level tech. I wish I could invest.

I also think they have the better solution to reenry. The spacex tiles give me shuttle flashbacks and I don't think they can be good enough for the mature vehicle.

 

Some thoughts I have, if anyone would like to discuss.

Is the aerospike/heat shield scalable? Could it scale to mars colonial fleet style, 20+ meter diameter monsters?

The heat shield requires cooling from cryofuel. Is anyone up for back of the napkin math on how much fuel they would need for this? With and without using that fuel to retroburn. Propulsive reentry gets complicated because it offset bow shock, reducing thermal conduction, and it widens bow shock, essentially acting like a grater diameter vehicle without the extra mass.

Do you guys think the high number of engines on the second stage are a hazard or a redundant safety. If a second stage engine failed, would symmetrically corresponding engines on the areospike ring need to be shutdown or throttled?

Edited March 13, 2023 by Superluminaut
units

[AckSed]

2 hours ago, Superluminaut said:

Is the aerospike/heat shield scalable? Could it scale to Mars colonial fleet style, 20+ foot diameter monsters?

We know that the wider a hypersonic blunt body becomes as it descends through the atmosphere, the lesser the heat load on the shield, as the compressive shock is pushed back. LOFTID reached this diameter, though it simply resisted the heat with silicon carbide fabric and was not carrying much load. An ablative heatshield helps resist heat by charring (creating a hard-to-burn outer layer) and vapourising (creating gas that carries away some of the heat and pushes back the shockwave).

Stoke's heatshield is not only actively cooled on the inside, it feeds out the exhaust gas through the centre, which, though it ignites, is still cooler than the re-entry plasma. The exhaust, still mostly hydrogen and a bit of water, then carries away some of the heat of the outside. The bulk of the heating in re-entry is radiative, so a shiny, cool, stainless steel surface will reflect some of that. Every little bit helps. It's also set up so that the hotter it becomes, the more H2 will flow.

So I'm mostly certain it will scale. In fact, I'm pretty sure that on a graph, the line of second stage wet mass versus hydrogen used for heatshield cooling would cease to be linear and begin to flatten out. This is only intuition though.

As we know, the limitation of an aerospike is its cooling. The fact that there are 15 small combustion chambers to one expander bleed cycle turbopump would also help wring useful work out of the heat.

3 hours ago, Superluminaut said:

The heat shield requires cooling from cryofuel. Is anyone up for back of the napkin math on how much fuel they would need for this? With and without using that fuel to retroburn. Propulsive reentry gets complicated because it offset bow shock, reducing thermal conduction, and it widens bow shock, essentially acting like a greater-diameter vehicle without the extra mass.

All I can give you is the strength of LH2 for cooling is not in its specific heat capacity (10 kilojoules per kilogram), but its heat of vapourisation (around 440 kJ per kg). I don't know the surface area of the heatshield. If we assume that it forms a spherical cap of a sphere... but it'd have to be a larger sphere to create the curve we saw.

I throw my hands up and ask someone who numbers gooder to calculate the area of a spherical cap based on where it intersects with the outer slightly-conical cylinder of the sidewalls and that of the upper stage, based on the depth of around 1 metre and an overall diameter of 4 metres.

4 hours ago, Superluminaut said:

Do you guys think the high number of engines on the second stage are a hazard or a redundant safety. If a second stage engine failed, would symmetrically corresponding engines on the aerospike ring need to be shutdown or throttled?

Mmm... not sure. There will be 30 combustion chambers, but one, maybe two turbopumps feeding them, so it's technically one or two engines. The fact that they are already aiming for thrust-vectoring by differential throttling would make shutting one combustion chamber down a minor inconvenience: they could close off the opposing chamber or just modulate the burn on the other chambers. If a turbopump failed, then that's bad. I don't know if they'll have two turbopumps feeding 15 chambers each, but I'd want redundancy. If they do, I'd say it's an asset.

[sevenperforce]

28 minutes ago, AckSed said:

If a turbopump failed, then that's bad. I don't know if they'll have two turbopumps feeding 15 chambers each, but I'd want redundancy. If they do, I'd say it's an asset.

The pumps are dual-redundant and plumbed to both sets of chambers:

Unless they are just running one turbpump for nothing, this means both turbopumps are running but supplying only half the chambers.

30 minutes ago, AckSed said:

I throw my hands up and ask someone who numbers gooder to calculate the area of a spherical cap based on where it intersects with the outer slightly-conical cylinder of the sidewalls and that of the upper stage, based on the depth of around 1 metre and an overall diameter of 4 metres.

What curves are you describing exactly? Because I can certainly do that but I'm not sure what you're referencing.

31 minutes ago, AckSed said:

Stoke's heatshield is not only actively cooled on the inside, it feeds out the exhaust gas through the centre, which, though it ignites, is still cooler than the re-entry plasma. The exhaust, still mostly hydrogen and a bit of water, then carries away some of the heat of the outside.

What's your source that the hydrogen is ignited during this process?

[AckSed]

7 minutes ago, sevenperforce said:

Unless they are just running one turbopump for nothing, this means both turbopumps are running but supplying only half the chambers.

Thanks.

8 minutes ago, sevenperforce said:

What curves are you describing exactly? Because I can certainly do that but I'm not sure what you're referencing.

Maybe I'm making it too complicated. I now realise what I want is the surface area of a dome 1 metre tall and 3.4 metres wide (I'm cutting out the outer circle with the ring of combustion chambers), the heat radiated by the plasma of the shockwave and the mass flow rate required to keep the surface of that area at around 200-500 deg. C.

...I am definitely not a rocket scientist.

14 minutes ago, sevenperforce said:

What's your source that the hydrogen is ignited during this process?

None, just assumed that the hydrogen would burn due to the heat once vented. I didn't mean that something would light it.

[sevenperforce]

32 minutes ago, AckSed said:

Maybe I'm making it too complicated. I now realise what I want is the surface area of a dome 1 metre tall and 3.4 metres wide (I'm cutting out the outer circle with the ring of combustion chambers), the heat radiated by the plasma of the shockwave and the mass flow rate required to keep the surface of that area at around 200-500 deg. C.

...I am definitely not a rocket scientist.

The area of a spherical cap with those dimensions is 18.22 m² and it's shaped like this:

That seems like the wrong shape. I'm thinking that the correct shape is probably something more akin to an ellipsoidal cap rather than a spherical cap. Probably closer to 16 square meters total.

Calculating the necessary coolant flow rate is going to take some effort. You're going to model the re-entry plasma as a 2000°C surface at constant temperature with an area equal to the heat shield. So you can just take that linearly, using the Stefan-Boltzmann law. The power per unit area coming off of a blackbody is given by j* = σT⁴, where σ is the Stefan-Boltzmann constant, 5.67e-8 W/m²K⁴ in Kelvin. So here the area-specific energy flux we're gonna need to deal with is 1.51 MW/m².

What's the heat shield made of, anyway? Stainless? Aluminum? I can't remember. Let's assume aluminum. The incoming energy is radiative, so you're going to need to look at the reflectivity of aluminum in the spectrum where the 2000°C Planck peak is located. The wavelength location of the Planck peak is given by Wien's displacement law, where λ = b/T and b is Wien's displacement constant (creative nomenclature here, I know) which is 2898 μm*K. So the plasma shell will have its radiative peak at ~1.27 micrometers which is solidly in the center of the infrared spectrum. Bare aluminum has a reflectivity of approximately 92% in infrared:

So that means the heat flux we actually have to deal with is only going to be 1.51 MW/m² x 8%, or 120.8 kW/m². Slightly more manageable. Now, before we bother with multiplying by the size of the heat shield, we also get to look at the blackbody emissions of the aluminum itself. Let's say we try to hold the heat shield at 500ºC, a solid 160 degrees lower than the melting point of aluminum. At 500°C or 773.15 K, a perfect blackbody radiates (according to the Stefan-Boltzmann law above) at 6117 W/m². Anodized aluminum has a blackbody emissivity of 77%, so the aluminum is going to reject about 4.71 kW per square meter. That's not much, but it's something, reducing our heat rejection needs to 116.1 kW/m².

(Note that the specifics of the metallic heat shield alloy are actually pretty important here. The more reflective you can make your material in the infrared, the better...but at the same time, it also benefits dramatically from a higher melting point because the Stefan-Boltzmann law makes radiative energy transfer go up with the 4th power of temperature. Stainless steel has a melting point about 2.7X as high as aluminum which translates to 42 times more blackbody energy transfer being shed before active cooling, assuming similar emissivity.)

While the hydrogen gas coolant bleed from the center of the heat shield (or elsewhere) will make the boundary layer larger and cause some additional expansion which will generally reduce coolant requirements, it will not directly absorb any of the radiation coming off of the plasma. Diatomic hydrogen is completely transparent to infrared light.

Assuming 16 square meters of heat shield need to be cooled, you're going to need to reject 1.86 MW of thermal energy during re-entry. If we just use the enthalpy of vaporization for liquid hydrogen, that would come to 4.23 kg of liquid hydrogen per second, slightly more than the hydrogen mass flow rate of an RL-10 engine at full throttle. However, that's worst-case scenario, and we aren't accounting for the fact that the hydrogen will also be expanded through a turbine, which converts a portion of that thermal energy into work (to pump itself).

The Stoke Space upper stage is comparable to the size of Crew Dragon, which has a plasma blackout of about 4-6 minutes. Now, the upper stage is much lighter than Crew Dragon and thus will slow down more quickly, but even using the numbers above and a 5-minute re-entry period, that would require the vaporization of just 1.27 tonnes of liquid hydrogen used up for active cooling, which is really quite modest.

[Superluminaut]

1 hour ago, sevenperforce said:

The area of a spherical cap with those dimensions is 18.22 m² and it's shaped like this:

That seems like the wrong shape. I'm thinking that the correct shape is probably something more akin to an ellipsoidal cap rather than a spherical cap. Probably closer to 16 square meters total.

Calculating the necessary coolant flow rate is going to take some effort. You're going to model the re-entry plasma as a 2000°C surface at constant temperature with an area equal to the heat shield. So you can just take that linearly, using the Stefan-Boltzmann law. The power per unit area coming off of a blackbody is given by j* = σT⁴, where σ is the Stefan-Boltzmann constant, 5.67e-8 W/m²K⁴ in Kelvin. So here the area-specific energy flux we're gonna need to deal with is 1.51 MW/m².

What's the heat shield made of, anyway? Stainless? Aluminum? I can't remember. Let's assume aluminum. The incoming energy is radiative, so you're going to need to look at the reflectivity of aluminum in the spectrum where the 2000°C Planck peak is located. The wavelength location of the Planck peak is given by Wien's displacement law, where λ = b/T and b is Wien's displacement constant (creative nomenclature here, I know) which is 2898 μm*K. So the plasma shell will have its radiative peak at ~1.27 micrometers which is solidly in the center of the infrared spectrum. Bare aluminum has a reflectivity of approximately 92% in infrared:

So that means the heat flux we actually have to deal with is only going to be 1.51 MW/m² x 8%, or 120.8 kW/m². Slightly more manageable. Now, before we bother with multiplying by the size of the heat shield, we also get to look at the blackbody emissions of the aluminum itself. Let's say we try to hold the heat shield at 500ºC, a solid 160 degrees lower than the melting point of aluminum. At 500°C or 773.15 K, a perfect blackbody radiates (according to the Stefan-Boltzmann law above) at 6117 W/m². Anodized aluminum has a blackbody emissivity of 77%, so the aluminum is going to reject about 4.71 kW per square meter. That's not much, but it's something, reducing our heat rejection needs to 116.1 kW/m².

(Note that the specifics of the metallic heat shield alloy are actually pretty important here. The more reflective you can make your material in the infrared, the better...but at the same time, it also benefits dramatically from a higher melting point because the Stefan-Boltzmann law makes radiative energy transfer go up with the 4th power of temperature. Stainless steel has a melting point about 2.7X as high as aluminum which translates to 42 times more blackbody energy transfer being shed before active cooling, assuming similar emissivity.)

While the hydrogen gas coolant bleed from the center of the heat shield (or elsewhere) will make the boundary layer larger and cause some additional expansion which will generally reduce coolant requirements, it will not directly absorb any of the radiation coming off of the plasma. Diatomic hydrogen is completely transparent to infrared light.

Assuming 16 square meters of heat shield need to be cooled, you're going to need to reject 1.86 MW of thermal energy during re-entry. If we just use the enthalpy of vaporization for liquid hydrogen, that would come to 4.23 kg of liquid hydrogen per second, slightly more than the hydrogen mass flow rate of an RL-10 engine at full throttle. However, that's worst-case scenario, and we aren't accounting for the fact that the hydrogen will also be expanded through a turbine, which converts a portion of that thermal energy into work (to pump itself).

The Stoke Space upper stage is comparable to the size of Crew Dragon, which has a plasma blackout of about 4-6 minutes. Now, the upper stage is much lighter than Crew Dragon and thus will slow down more quickly, but even using the numbers above and a 5-minute re-entry period, that would require the vaporization of just 1.27 tonnes of liquid hydrogen used up for active cooling, which is really quite modest.

I just did a quick pass over this and don't fully grasp it yet. But I think reflectivity is not significant here. I believe the heating occurs through conduction/convection, not radiation.

A simple thought experiment seems to confirm my assumption. If radiation was a significant heat source, the shuttle tiles would never have been black.

[tater]

Heat shield is steel.

[sevenperforce]

4 hours ago, tater said:

Heat shield is steel.

Thanks, that changes the math somewhat. Source?

5 hours ago, Superluminaut said:

I just did a quick pass over this and don't fully grasp it yet.

I can understand, lol. It's a lot of work and people unfamiliar with high-temperature physics won't get it immediately.

5 hours ago, Superluminaut said:

I think reflectivity is not significant here. I believe the heating occurs through conduction/convection, not radiation.

You are incorrect. Initially, there is convective and conductive heating as molecules bounce off of the surface at high speeds. The extreme heating of the SR-71 windscreen, for example, is wholly the result of convective and conductive heating. However, if that's all you had to deal with, you could simply use a heat shield with low thermal conductivity and thus eliminate the heat transfer completely.

The amount of energy transferred to the vehicle from the surrounding medium by conductive and convective heating goes up with the third power of vehicle velocity; the amount of energy transferred to the vehicle from the surrounding medium by radiative heating goes up with the eighth power of vehicle velocity. So when you're moving at re-entry speeds, radiative heating dominates.

5 hours ago, Superluminaut said:

A simple thought experiment seems to confirm my assumption. If radiation was a significant heat source, the shuttle tiles would never have been black.

Sometimes intuitions based on simple thought experiments are correct. This is not one of those times. The Fibrous Refractory Composite Insulation (FRCI) tiles on the Shuttle were black because of that T⁴ term in the Stefan-Boltzmann law. Even though the color black is not as reflective as colors like white and silver, black is a much more efficient emitter of thermal radiation than white or silver, and so the Shuttle tiles functioned by accepting as much heat as the plasma could produce and then just re-radiating that heat away from the vehicle.

Ablative heat shields do quite a bit of radiating, but they also dump a lot of heat by allowing the oils impregnated into their carbon structure to burn away and carry that absorbed radiative heat away from the vehicle.

For an actively-cooled low-temperature heat shield like the one on Stoke's vehicle, the shield never gets very hot and so the emissivity of the shield isn't as meaningful because the Stefan-Boltzmann law doesn't give as much of an advantage. 

[tater]

41 minutes ago, sevenperforce said:

Thanks, that changes the math somewhat. Source?

He says so directly in the Tim Dodd interview.

[tater]

I double-checked, and he doesn't say the heat shield is steel explicitly that I found, but he talks about them choosing steel as the material for the vehicle to optimize for manufacturing in the section that is about the heat shield when they walk around the vehicle, then visit the heat shield. He talks about steel keeping properties at higher temps as well.

Edited March 13, 2023 by tater

[magnemoe]

2 hours ago, tater said:

I double-checked, and he doesn't say the heat shield is steel explicitly that I found, but he talks about them choosing steel as the material for the vehicle to optimize for manufacturing in the section that is about the heat shield when they walk around the vehicle, then visit the heat shield. He talks about steel keeping properties at higher temps as well.

Then you want steel for the heat shield as it will get hot, You might use lighter materials for the cargo hold and nose but pretty sure the heat shield and probably the tanks are steel. 

[tater]

1 hour ago, magnemoe said:

Then you want steel for the heat shield as it will get hot, You might use lighter materials for the cargo hold and nose but pretty sure the heat shield and probably the tanks are steel. 

I got the impression it's all steel, but yeah.

[Superluminaut]

20 hours ago, sevenperforce said:

You are incorrect. Initially, there is convective and conductive heating as molecules bounce off of the surface at high speeds. The extreme heating of the SR-71 windscreen, for example, is wholly the result of convective and conductive heating. However, if that's all you had to deal with, you could simply use a heat shield with low thermal conductivity and thus eliminate the heat transfer completely.

I don't think this is a useful comparison. The blackbird was a dart and lacked the insulating boundary layer that blunt bodies produce between the bows shock wave and vehicle.

And I think the shuttle tiles were just that, painted insulation. Most of it being a silicone styrofoam like brick, if I remember right.

19 hours ago, sevenperforce said:

The amount of energy transferred to the vehicle from the surrounding medium by conductive and convective heating goes up with the third power of vehicle velocity; the amount of energy transferred to the vehicle from the surrounding medium by radiative heating goes up with the eighth power of vehicle velocity. So when you're moving at re-entry speeds, radiative heating dominates.

Sometimes intuitions based on simple thought experiments are correct. This is not one of those times. The Fibrous Refractory Composite Insulation (FRCI) tiles on the Shuttle were black because of that T⁴ term in the Stefan-Boltzmann law. Even though the color black is not as reflective as colors like white and silver, black is a much more efficient emitter of thermal radiation than white or silver, and so the Shuttle tiles functioned by accepting as much heat as the plasma could produce and then just re-radiating that heat away from the vehicle.

I still think the shuttle tile can be used as a reality check on the math.

I've done a bit of reflectivity testing in the past and an artificial black can easily dip under 0.1 reflectivity, while white can top 0.9. So reflectivity limitations can be ruled out as a  factor in color choice.

Therefor, if radiative heating was the bulk source of heat, the shuttle tiles should have all been white.  It's better to reject 90% of energy than it is to absorb it and then reject it.

Also only the leading edge of the shuttle had black tiles, yet plasma is fluid. It would flow around the vehicle and radiate it from all angles. However white was the choice for the trailing end of the vehicle, so black must have been selected for another reason.

Surely at a higher speed that 8th power term would make radiative heating dominant, but it does not appear to be the case for LEO for whatever reason. Possibly the actual area of plasma producing that amount of radiation is very small.

Even during reentering from the moon astronauts had windows and did not wear eye or skin protection, so any plasma wrapping around the vehicle could not have been particularly bright.

[sevenperforce]

2 hours ago, Superluminaut said:

On 3/13/2023 at 6:54 PM, sevenperforce said:

The extreme heating of the SR-71 windscreen, for example, is wholly the result of convective and conductive heating. However, if that's all you had to deal with, you could simply use a heat shield with low thermal conductivity and thus eliminate the heat transfer completely.

I don't think this is a useful comparison. The blackbird was a dart and lacked the insulating boundary layer that blunt bodies produce between the bows shock wave and vehicle.

That was my point.

2 hours ago, Superluminaut said:

I still think the shuttle tile can be used as a reality check on the math.

Not unless you have the correct math.

2 hours ago, Superluminaut said:

I've done a bit of reflectivity testing in the past and an artificial black can easily dip under 0.1 reflectivity, while white can top 0.9. So reflectivity limitations can be ruled out as a  factor in color choice.

What spectral bands was your reflectivity testing in? Infrared light is not visible light. Many things that are opaque to visible light are transparent to infrared. Although the grey-black tiles on the bottom of the Shuttle absorbed the majority of visible light, they were reasonably reflective of infrared radiation.

2 hours ago, Superluminaut said:

Therefor, if radiative heating was the bulk source of heat, the shuttle tiles should have all been white.  It's better to reject 90% of energy than it is to absorb it and then reject it.

You're not doing the math.

Imagine your heat shield needs to get rid of 98% of the incoming energy. If you have a white heat shield which reflects 90% of incoming energy but can only radiate away 50% of what it absorbs, it will get rid of 95% of incoming energy, which means it fails. If you have a black heat shield which only reflects 75% of incoming energy but can radiate away 93% of what it absorbs, then it will get rid of 98.25% of the incoming energy, so it survives.

2 hours ago, Superluminaut said:

Also only the leading edge of the shuttle had black tiles, yet plasma is fluid.

The leading edges of the wings had black carbon-carbon composite; the entire underside of the orbiter was black tiles.

2 hours ago, Superluminaut said:

It would flow around the vehicle and radiate it from all angles.

It did. That's why you need a thermal protection system on the leeward side of a re-entry vehicle. Look, for example, at the metallic reflective heat shield on the lee side of Orion.

However, as plasma starts to flow around a vehicle, it expands. Remember that the reason plasma is plasma in the first place is that it is air compressed by the vehicle's high velocity; the air literally cannot get out of the way fast enough. Once it starts to expand, it isn't nearly so hot, and so you can have a backshell thermal protection system which can function by reflecting heat and doesn't need to radiate away heat at all.

2 hours ago, Superluminaut said:

However white was the choice for the trailing end of the vehicle, so black must have been selected for another reason.

It was. Black was chosen despite being less reflective, because it cools itself radiatively much better than a polished reflective surface.

2 hours ago, Superluminaut said:

Surely at a higher speed that 8th power term would make radiative heating dominant, but it does not appear to be the case for LEO for whatever reason.

It does, in fact, appear to be the case in LEO. And it is, in fact, the case in LEO.

2 hours ago, Superluminaut said:

Possibly the actual area of plasma producing that amount of radiation is very small.

The actual area of plasma producing that amount of radiation is very large. That's why you can see shooting stars.

2 hours ago, Superluminaut said:

Even during reentering from the moon astronauts had windows and did not wear eye or skin protection, so any plasma wrapping around the vehicle could not have been particularly bright.

 Definitely no eye or skin protection visible in this photo:

Spoiler

 

[RyanRising]

On 3/14/2023 at 5:13 PM, sevenperforce said:

It does, in fact, appear to be the case in LEO. And it is, in fact, the case in LEO.

The actual area of plasma producing that amount of radiation is very large. That's why you can see shooting stars

I think you’ve got things mixed around. It’s worth noting that while you see silver reflective back shells on Orion and Apollo, you do not see those on LEO spacecraft, which should tell you about the different types of heating they experience.

But if we just make numbers up and only look ant the visible features of spacecraft, even if we do math on those made-up numbers, we’d be here all day. Instead, let’s look at some actual material on the topic. Here’s what Mr. H. Julian Allen of NASA Ames has to say about the topic:

on ballistic missiles entering from near-orbital velocity: 

Quote

Now, the heating rate for these vehicles is very large compared to the rate at which heat can be reradiated from the surface, and the driving temperature potential promoting the convective transfer of heat is determined essentially by the air temperature (i.e., the wall temperature can be ignored by comparison to the air temperature).

Of manned space vehicles at orbital velocity:

Quote

In contrast, for manned vehicles which must employ shallow-angled
trajectories to avoid excessive accelerations, the heating rate is more modest but lasts for a considerable period. The conductivity problem is so severe that insulation is required to prevent overheating of the substructure.
…

Also for large heavy vehicles the flight Reynolds number is lower so that a laminar flow can be maintained where otherwise turbulent flow would occur - hence, the heating rate is lessened.

This implies that convective effects make up a significant amount of the heat transfer for atmospheric entry vehicles. However, what really seals the deal is this little transition:

Quote

Up  to this point the tacit assumption has been made that convective
heating constitutes the total. For speeds up t o nearly parabolic speed
for Earth, this is essentially the case.

He outright says that the heating is still essentially convective at speeds even exceeding that of LEO. At this point, the Mercury orbital flights had already been performed, so if NASA had designed its TPS, and that of Gemini, assuming this to be true when in reality radiative heating dominated the entry environment, they would have cooked their astronauts. 
Mr. Allen then goes on to describe the radiative heating effects that occur with the higher-energy plasmas associated with super orbital velocities, and yes, this was clearly important enough that a reflective backshell had to be incorporated onto Apollo where the previous spacecrafts‘ leeward sides were designed to essentially be black bodies.

https://ntrs.nasa.gov/api/citations/19640013352/downloads/19640013352.pdf

And from another report, https://ntrs.nasa.gov/api/citations/19980227977/downloads/19980227977.pdf, here’s a handy graph (though in imperial units, bleh) of the rough relationship of convective and radiative heating. As you can see, in the regime of typical LEO spacecraft entries the convective heating is greater than the radiative by some orders of magnitude.

[RyanRising]

Radiative heat transfer from the vehicle is indeed very significant during shallow entries, though. That is indeed why the high-temp shuttle tiles were black, and wouldn’t have worked well at all if the majority of the heat coming into them was radiative. 

[AckSed]

So what kind of heat load would the heatshield experience upon re-entry from LEO? Are we talking 'rocket combustion chamber' temperatures and pressures? Higher?

[sevenperforce]

 

32 minutes ago, AckSed said:

So what kind of heat load would the heatshield experience upon re-entry from LEO? Are we talking 'rocket combustion chamber' temperatures and pressures? Higher?

Comparable temperatures, but much lower pressures. However, rocket combustion chambers employ active cooling, which heat shields (up until now) haven't had.

Of course, you can consider ablative cooling to be a form of active cooling.

1 hour ago, RyanRising said:

Radiative heat transfer from the vehicle is indeed very significant during shallow entries, though. That is indeed why the high-temp shuttle tiles were black, and wouldn’t have worked well at all if the majority of the heat coming into them was radiative. 

That was my primary point: that heat shield tiles were black because it was necessary that they radiatively cool themselves.

2 hours ago, RyanRising said:

And from another report, https://ntrs.nasa.gov/api/citations/19980227977/downloads/19980227977.pdf, here’s a handy graph (though in imperial units, bleh) of the rough relationship of convective and radiative heating. As you can see, in the regime of typical LEO spacecraft entries the convective heating is greater than the radiative by some orders of magnitude.

I think you're making an assumption here about what the "regime of typical LEO spacecraft entries" constitutes. It's not an either/or distinction between radiative and convective heating. Rather, re-entries start with radiative heating as the dominant factor, and then the radiative heating decreases while the convective heating increases.

You're correct that certain vehicles are designed to accept convective heat and then radiate it away. However, the image you provided above (Figure 10) isn't applicable to LEO entries generally. The source you cite says they "apply only to single-pass, nonlifting, parabolic-velocity entries." The real numbers are more complicated.

The actual numbers can be found in this presentation, which explains the relationship between convective heating and radiative heating based on multiple factors. Importantly, one of the factors is the effective radius of the vehicle:

Q_conv ∝ v³(ρ/R)^0.5 but Q_rad ∝ v⁸ρ^1.2R^0.5

As the effective vehicle radius increases, convective heating decreases, but radiation heating increases. For something like the Shuttle, which had a very deep atmospheric entry (to allow a lifting entry) and a very high effective radius, it would make sense that radiative heating was a major component.

[AckSed]

3 hours ago, sevenperforce said:

Comparable temperatures, but much lower pressures. However, rocket combustion chambers employ active cooling, which heat shields (up until now) haven't had.

How would water-cooled heat shields (which I've seen mentioned here and there) stack up against H2-cooled? I imagine they have a denser coolant and pretty damn good cooling, but the system is dead weight that doesn't contribute to propulsion.