# ApolloHoax.net

## Apollo Discussions => The Reality of Apollo => Topic started by: Allan F on April 26, 2013, 05:30:14 PM

Title: Is my math correct? LM descent engine force at nozzle
Post by: Allan F on April 26, 2013, 05:30:14 PM
I was wondering, what the force of the exhaust of the LM's descent engine was, when the LM was close to landing, and the descent fuel was almost used up.

So I used the formula F = m x a, with mass m 6.800 kg, acceleration a 1.62 m/s, to calculate what the force needed to balance the lunar gravity. and got 11000 N. That's the equivalent of lifting 1120 kg on earth. So far so good.

The nozzle area is pi x r^2, with r being the radius of the nozzle at the bottom - 0.75 meters. That's 1.766 m^2. That's 17660 cm^2.

That gives a force at the nozzle of 0.62 N/cm^2.

This force seems very very low, so I would like a reality check on my math, please.

Edit: That equates to a force equivalent to 62 grams/cm^2.
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: ka9q on April 26, 2013, 10:15:52 PM
Yeah, your calculation sounds about right. But I'm not sure what the result actually means.

I'm still learning all this stuff myself, so maybe you can help me figure it out. The nozzle is a heat engine that turns the random thermal motions of the slow moving, hot, high pressure gas molecules in the combustion chamber into a cool, low pressure but rapidly moving exhaust gas. Ideally all of the heat is turned into kinetic energy and you have a rapidly moving exhaust stream at zero temperature and zero pressure.

You could do this in a vacuum with an infinitely long nozzle but that's not practical. The conversion can't be complete. As in any heat engine, some of the input heat is wasted as residual heat at a lower temperature. The exhaust is much cooler than in the combustion chamber, but still hot. So it has residual pressure, and in a vacuum the plume will immediately start to expand sideways as it leaves the nozzle, and those molecules don't contribute to thrust. Depending on why you're calculating it (e.g., to examine the effects on the lunar surface) you might want to take it into account.

The textbooks say that rocket engines produce two kinds of thrust: momentum thrust and pressure thrust. Momentum thrust is easy to understand; it's the actual exhaust velocity times the mass flow rate. Classic action-reaction. Momentum thrust is always positive, but it depends on the efficiency of the nozzle in converting heat to kinetic energy. It's generated when molecules bounce off the top of the combustion chamber, or the inside of the nozzle, transfer their momentum, and then leave without hitting any other part of the engine.

I'm still trying to grok pressure thrust. It's defined as the nozzle exit area times the difference between exhaust and ambient pressure. In an atmosphere this can be either positive or negative, but in a vacuum it's always positive. Optimum efficiency occurs when the exhaust pressure is exactly equal to ambient and pressure thrust is zero, because this maximizes both momentum thrust and overall thrust. There's an ideal nozzle length in an atmosphere. In a vacuum, a longer nozzle is always better (except for weight and size).

What bothers me is that "pressure thrust" sounds a lot like the misconception that rockets work by pushing on the air. Obviously that's not the case here, but I don't fully understand it.

It's easy to see that the random heat motions that happen to be sideways don't contribute to thrust. What I don't quite understand is the effect of those residual thermal motions that happen to be in the desired direction, i.e., parallel to the exhaust flow, but I think this is where pressure thrust comes from. I'm trying to visualize how these molecules transfer their momentum to the rocket. I think it's indirect; they bounce into each other until one eventually bounces into the engine nozzle.

What I think you've computed is yet another kind of pressure: stagnation pressure. If I understand it correctly, it's what you get when the plume impacts a large flat surface. Although the pressure in the plume may be very low (i.e., it doesn't tend to expand), when it hits the molecules will deliver their momentum in the process of being deflected sideways. Many will collide with each other and the plume will heat up as well.
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: Allan F on April 26, 2013, 10:37:56 PM
Yes, that seems to be the stagnation pressure. I'm aware of the sideways expansion, once the exhaust leaves the nozzle. And you're correct, it is to get a feel for the impact it had on the surface.

I don't think the exhaust has much effect on the force on the spacecraft, once the gas has exited the nozzle.

And I seem to have forgot a "^2" after 1.6 m/s.

Edit: The rest of your post is beyond my current level, but I'll see if I can find some reading material to improve my understanding.
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: ka9q on April 26, 2013, 11:12:30 PM
For what it's worth, I've been playing with the equations for isentropic flow in an ideal rocket nozzle, so here are the numbers I calculated for the LM DPS at 20% throttle, which is about where it was at the Apollo 11 landing.

The chamber pressure in psi was almost exactly equal to the throttle setting in percent, so the chamber was 20 psi.

The expansion ratio is given as 47.5:1 and the exit diameter as 59.0 inch. From that I compute a throat area of  0.0372 m^2. (Caveat: the J-series LMs had longer nozzles, and I'm not sure where this 47.5:1 figure came from. I think it's for the pre-J LMs, including Apollo 11).

From Bob B's web pages I extracted the following information for Aerozine-50 & N2O4:

k = 1.228
mol wt = 21.8 g/mol
adiabatic flame temp = 3100 K

I may have done this wrong; these values are from graphs that vary with chamber pressure and mixture ratio, and I didn't know the latter offhand. I had to extrapolate his plots down to the LM's chamber pressure, and Bob himself gave the caveat that, unlike other propellants, his calculations seem to differ from practice for AZ-50/N2O4.

Anyway, I compute the following:

Mass rate = 3.09 kg/s
exit temp = 927 K
exit press = 207 Pa (.002 atm!; 25-50% Mars atmospheric pressure)
exit density = .000585 kg/m^3 (0.05% of sea level air)
exit velocity = 2988 m/s
Isp = 317 sec, effective velocity = 3106 m/s
Carnot efficiency = 70.1% (efficiency of nozzle as heat engine = 1 - exit_temp/chamber_temp)
Total thrust = 9590 N
Momentum thrust = 9225 N
Pressure thrust = 365 N

As a check, the Isp is specified to be 311 s at an unknown throttle setting, and actual landing thrust would be 20% * 45.04 kN = 9 kN. So I'm not too far off given the questionability of some of my source numbers. But it's rather striking to see the extremely low density, pressure and temperature of the exhaust, and of course these figures would be even lower by the time the plume impacted the surface.

Oh, and these low numbers underscore just how silly Ralph Rene's leaf blower "tests" really were.

Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: ka9q on April 26, 2013, 11:27:38 PM
Edit: The rest of your post is beyond my current level, but I'll see if I can find some reading material to improve my understanding.
It was beyond mine too until just recently, when I decided to finally try to learn the thermodynamics of rocket engines. I was prompted by an argument with yet another Youtube moron who claimed that rockets can't work in a vacuum because they push on air. As always, I learn something while they learn nothing.
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: Allan F on April 26, 2013, 11:53:01 PM
Your molar weight seem low. Nitrogen tetroxide should be 4x16 + 2x14 = 92 g/mol and aerozine 50, (60 + 32)/2 = 46.

Yes, the gain seems to be one-sided in these discussions.

Edit: The molar weight you quote, is that for the combustion products?
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: Allan F on April 26, 2013, 11:58:54 PM
What numbers do you use for the nozzle? I've been using 1.5 meter, but I'm willing to accept a superior source.
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: ka9q on April 27, 2013, 12:24:45 AM
Edit: The molar weight you quote, is that for the combustion products?
Yes, it's the combustion products that count. For AZ-50 & N2O4 at the usual rich mixture I'd expect mostly CO, CO2, N2, H2O and H2.

What complicates the process is that the equilibrium among these compounds changes as the exhaust cools in the nozzle. This is called "shifting equilibrium" but it's currently beyond my level to analyze.
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: ka9q on April 27, 2013, 12:47:37 AM
What numbers do you use for the nozzle? I've been using 1.5 meter, but I'm willing to accept a superior source
I don't see a nozzle or throat diameter specifically listed, but as with the F-1 and J-2 engines I took the advertised engine diameter and assumed that was the diameter of the nozzle exit (as it appears to be in the pictures). The figure I found for the LM is 59.0 inch or 150 cm, but I don't know for sure if that's the pre-J or J version.
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: VQ on April 27, 2013, 01:11:30 AM
0.62 N/cm^2 isn't all that low of pressure, either. That is 6200 N/m^2=6200 Pa, or about 6% of atmospheric pressure at sea level.
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: ka9q on April 27, 2013, 01:12:33 AM
Another interesting thing I've noticed is if you compute an "equivalent thrust area" by dividing the total thrust by the chamber pressure, you get an area somewhere between 1.5 and 2 times the throat area. I take this to mean that somewhere between half and two-thirds of the total thrust is generated by hot, high pressure gas pushing on the top of the combustion chamber opposite the nozzle, and the remainder is generated by the expanding, lower pressure gas pressing on the inside of the much larger nozzle as it expands. You can visualize those molecules producing thrust as they bounce off the curved shape of the nozzle, deposit their momentum and are redirected toward the exit.
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: ka9q on April 27, 2013, 01:15:32 AM
0.62 N/cm^2 isn't all that low of pressure, either. That is 6200 N/m^2=6200 Pa, or about 6% of atmospheric pressure at sea level.
It's interesting to compare that to my calculated 207 Pa exit pressure. That's what you'd read on a gauge drilled through the edge of the nozzle right before the gas exits. Your 6200 Pa reflects the "stagnation" pressure, the total pressure you'd get as the fast-moving gas slams into a blunt obstacle.

The engine does a pretty good job of turning all that heat and pressure into kinetic energy, but you can turn it back to heat and pressure by putting a solid barrier in the flow.
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: AtomicDog on April 27, 2013, 12:18:50 PM
Eh, it's all Rocket Science to me.
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: Daniel Dravot on April 27, 2013, 01:18:58 PM
I was prompted by an argument with yet another Youtube moron who claimed that rockets can't work in a vacuum because they push on air.

Seems an awful waste, carrying all that oxygen around, when the damn things only work in places where oxygen is available for free.
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: ka9q on April 27, 2013, 02:17:49 PM
Yeah, I keep pointing that out. The Saturn V was something like 2/3 LOX by weight at launch. They never seem to have an answer for that. In fact, they never seem to have an answer for a lot of questions that would back them into a corner.
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: Allan F on April 27, 2013, 04:03:52 PM
So the throat of the engine was about 11 cm in diameter?

Funny, how details after details keep popping up. Once you get into the workings, layer upon layer of complexity reveals itself. That is also one of the problems the conspiracists struggle with. The inability to grasp that complexity, that somebody actually has worked all these details out.
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: ka9q on April 27, 2013, 04:45:22 PM
The DPS engine diameter is given as 59 inches, and since the nozzle seems to be the widest part I assume that's the diameter of the nozzle exit. That corresponds to an exit area of 1.76 m^2.

The expansion ratio, which always refers to the areas, is given as 47.5:1. (I don't know offhand if that's for the J or pre-J models.) That implies a throat area of 1.76 / 47.5 = .03713 m^2. That in turn corresponds to a throat diameter of 21.74 cm. Your 11 cm would be the radius of the throat, half the diameter.
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: Allan F on April 27, 2013, 04:51:41 PM
Math in the head. Not always the best way.

Seeing into the engine bell on A17's ascent, the 11 cm seemed a little on the narrow side. 22 is much better.

The J-missions 15-16-17, did they have a different/extended nozzle? Maybe to compensate for the extra load of the LRV and consumables?
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: ka9q on April 28, 2013, 02:48:02 AM
The J missions had a longer nozzle on the descent engine to handle the larger load. The longer nozzle increases the expansion ratio and the efficiency of the nozzle in turning heat energy into kinetic energy.

Apollo 15's DPS nozzle was crunched on landing, though this was not unexpected.
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: smartcooky on April 28, 2013, 03:04:20 AM
Yeah, I keep pointing that out. The Saturn V was something like 2/3 LOX by weight at launch. They never seem to have an answer for that. In fact, they never seem to have an answer for a lot of questions that would back them into a corner.

Oh yes they do....flouncey flouncey!!

Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: Daniel Dravot on April 28, 2013, 06:50:00 AM
Yeah, I keep pointing that out. The Saturn V was something like 2/3 LOX by weight at launch. They never seem to have an answer for that. In fact, they never seem to have an answer for a lot of questions that would back them into a corner.

When they talk about rockets pushing against the air, it causes me to imagine one with something that looks like a steamboat paddle attached to the back.

Maybe if I have some time, I'll do a photoshop :)
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: Donnie B. on April 28, 2013, 08:33:25 AM
Yeah, I keep pointing that out. The Saturn V was something like 2/3 LOX by weight at launch. They never seem to have an answer for that. In fact, they never seem to have an answer for a lot of questions that would back them into a corner.

When they talk about rockets pushing against the air, it causes me to imagine one with something that looks like a steamboat paddle attached to the back.

Maybe if I have some time, I'll do a photoshop :)

I like that idea.  The Robert J. Fulton rocket.

It's absurd on the face of it.  What is that stream of hot gas supposed to push against?  Are the atmospheric nitrogen and oxygen molecules supposed to just hang around and push back?  Or solidify into a sturdy column as the rocket rises?  It makes you wonder if these lackwits have ever heard of something called "wind".  Come to think of it, the only way their concept would work was if the rocket engines were mounted on the ground firing upward, so as to blow the rocket into space...

We have technology that "pushes against the air" -- it's called a propeller.  Can you imagine the size and/or number of propellers required to get a Saturn V off the ground -- without even the benefit of wings?
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: Noldi400 on April 28, 2013, 12:34:54 PM
Math in the head. Not always the best way.

Seeing into the engine bell on A17's ascent, the 11 cm seemed a little on the narrow side. 22 is much better.

The J-missions 15-16-17, did they have a different/extended nozzle? Maybe to compensate for the extra load of the LRV and consumables?

Not having much of a head for physics (biology fits much better), I'm way behind both of you, but NASA has a page that I've found helpful:

http://www.grc.nasa.gov/WWW/k-12/airplane/lrockth.html (http://www.grc.nasa.gov/WWW/k-12/airplane/lrockth.html)

It has a link to a nozzle simulator applet where you can change the nozzle parameters and see the effect it has on thrust and other factors.

Speaking of HBs who think a rocket pushes against the air, it seems to me that another common characteristic of a lot of them is a preference for believing what their "common sense" or what they think of as "intellect" tells them rather than doing some real-world research.

The question of "why don't you hear engine sounds on the voice recordings" comes to mind...  they never seem willing to take the simple step of listening to voice tapes of other people in the cockpits of aircraft with noisy engines, they just stick with their preconceptions.  Or the old "blast crater" chestnut - insisting that there should be one despite the fact that Harriers, with a much higher thrust/cm2 than the LM, don't make craters when landing on dirt surfaces.

Or - one of my personal favorites - Hunchbacked's insistence that the LM should have landed by coming to a complete stop horizontally, then descending vertically, even though any pilot of a helicopter or other VTOL craft will tell you that (1) if at all possible they want to be moving forward during landing (you can't see what's directly under you)  and (2) holding a hover with no horizontal movement is one of the hardest things they have to do.  Not to mention that hovering with a rocket engine is the most horrendously fuel-wasting thing you can do.

Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: smartcooky on April 28, 2013, 03:54:28 PM
Or - one of my personal favorites - Hunchbacked's insistence that the LM should have landed by coming to a complete stop horizontally, then descending vertically, even though any pilot of a helicopter or other VTOL craft will tell you that (1) if at all possible they want to be moving forward during landing (you can't see what's directly under you)  and (2) holding a hover with no horizontal movement is one of the hardest things they have to do.  Not to mention that hovering with a rocket engine is the most horrendously fuel-wasting thing you can do.

I can vouch for that. I have tried doing this in a real helicopter; a Bell 47G (for those who don't know this helicopter, think M.A.S.H.).

Picture this.

You have your left hand on a lever that is rather like a handbrake between the bucket seats of a small car. This is the "collective" lever; lifting it increases the pitch of the main rotor. On this lever is a twist grip like a motorbike throttle. Twisting it to the left increases the rpm of engine.

You have your feet on two pedals on the floor in a similar position to the pedals on your car. They are connected to the tail rotor and change its pitch. pushing on the left pedal pushes the tail to the left (therefore rotating the helicopter clockwise). Vice versa for the right.

Finally, your right hand is on the control stick, Pushing the control stick on any direction tilts the helicopter in that direction.

Now, even in dead-calm conditions, the hovering chopper will drift in a direction, say, left. When you try to correct it by pushing the control column right, if you make the slightest input forward or back when trying to centralise the control column, it will start to drift in that direction too. Also, when the helicopter drifts, it begins to lose altitude, because the down-draft is no longer pushing directly downwards, so you have to lift up the collective lever slightly and increase the throttle a notch, but doing that will cause the helicopter to counter rotate against the torque of the increased main rotor speed, so you have compensate for that with the pedals. As the helicopter comes upright, the increased main rotor pitch and speed causes the helicopter to climb, so you need to drop the collective and rpms a little, which of course, affects the counter rotation.

Add to that the gyroscopic effects of the main rotor, and you can see that hovering a helicopter is no easy task.

...and this is what will happen if you don't know how all this works...

Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: Allan F on April 28, 2013, 03:57:50 PM
How did he get in there without an instructor? No way he was ready to solo.

Edit: My old highschool had work done on it's roof. 4 large aircondition/ventilation units, size of a small van, had to be lifted and placed on the roof. Due to shape of the landscape, no crane was able to reach up and in to put them in place. Maersk Air provided a helicopter (for a price), and the pilots and crew put the gear in place with milimetre precision. I stood on the roof, a couple of dozen meters away, and was quite amazed by the precision they were capable of. I could see into the cockpit, and could clearly see the pilot's minute, barely visible, control inputs on the cyclic.
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: smartcooky on April 28, 2013, 04:20:58 PM
How did he get in there without an instructor? No way he was ready to solo.

Edit: My old highschool had work done on it's roof. 4 large aircondition/ventilation units, size of a small van, had to be lifted and placed on the roof. Due to shape of the landscape, no crane was able to reach up and in to put them in place. Maersk Air provided a helicopter (for a price), and the pilots and crew put the gear in place with milimetre precision. I stood on the roof, a couple of dozen meters away, and was quite amazed by the precision they were capable of. I could see into the cockpit, and could clearly see the pilot's minute, barely visible, control inputs on the cyclic.

Yep, in flying a helicopter, there really is there is no substitute for experience.
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: Allan F on April 28, 2013, 05:03:44 PM
It is all reaction programming. One's reflexes are 3-4 maybe even 10 times faster than deliberate conscious decisions. That's why a secretary can type 600 letters/minute, and others use the biblical approach on the keyboard. That's why those who practise martial arts are able to deflect and throw bigger and stronger opponents, race car drivers can do extreme things with cars.
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: Allan F on April 28, 2013, 08:13:37 PM
Talking about rockets and air, there's something I have been wondering about. That is: Is it possible to improve the efficiency of a rocket engine, firing in atmosphere, by adding a ring around the nozzle? In a way, so there is room between the ring and the nozzle, and the ring extends from somewhere above to somewhere below the nozzle. I have nothing to back this up, but my theory is, that the exhaust will pull ambient air into the direction, increasing the mass flow of the exhaust. Somewhat similiar to the way a propeller on a boat is more efficient, if it is ducted?
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: Daniel Dravot on April 28, 2013, 10:58:15 PM
...and this is what will happen if you don't know how all this works...

Ouch.

So if I am interpreting what I am seeing correctly, the pilot-in-training emerges from the wreckage and stalks off rather sheepishly.

I'm surprised he didn't realise he was in trouble while still skipping across the ground, and cut power - the outcome might not have been so bad in that case.

He did manage to provide some free lift for the wing of that craft in the foreground at the end.
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: Allan F on April 28, 2013, 11:17:06 PM
Once he was off the ground, the smart thing (I suspect) would be to gain altitude, sort himself out, get stable, and then work from there.

It looked like he had his left and right pedal confused. And once the tail rotor touched the ground, all control was lost.
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: ka9q on April 29, 2013, 02:05:42 AM
I have nothing to back this up, but my theory is, that the exhaust will pull ambient air into the direction, increasing the mass flow of the exhaust. Somewhat similiar to the way a propeller on a boat is more efficient, if it is ducted?
I don't think so, because it wouldn't affect the forces on the inside of the rocket engine where all the thrust is generated. It might have some second-order effects on the pressures around the base of the rocket, but offhand I can't say what they'd be. It might just add drag.

If you used the ring to extend the nozzle, this would increase performance only if the ring were flared to permit additional expansion, and only if the nozzle isn't already overexpanded. A nozzle is never overexpanded in vacuum, but in air it will be overexpanded if the exit pressure is below ambient.

Some engines have nozzle extensions that can be deployed in flight, and they look much like your ring except that it's flared and moves. It's launched with the extension retracted and during flight it is lowered into place with jack screws until it contacts the nozzle and effectively makes it longer.

This would be used on a ground-lit engine to allow it to avoid overexpansion on the ground while increasing the expansion ratio after reaching altitude to reduce underexpansion and improve performance. I don't know how well it works.
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: ka9q on April 29, 2013, 02:19:24 AM
Picture this.

To this non-pilot engineer, this seems like an application crying out for modern control technology. Use an IMU (inertial measurement unit) plus GPS in a closed-loop control system to set the throttle, collective and rotor tilt so as to provide whatever state (orientation, position and velocity) the pilot wants as indicated by pair of joysticks. Or an autopilot.

No doubt the purists would consider this "cheating", but I point out that our favorite helicopter-like flying machine, the Apollo LM, worked in much this way. Only the specific mechanisms for producing the reactive forces on the vehicle were different.

I understand that not all states are immediately accessible, e.g., you cannot start translating from a hover without changing your attitude at least momentarily so the pilot would still have to take this into account.

Don't those small hobby helicopters and other flying machines already have systems like this? If so, why not the big machines too?

Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: Daniel Dravot on April 29, 2013, 02:44:54 AM
Once he was off the ground, the smart thing (I suspect) would be to gain altitude, sort himself out, get stable, and then work from there.

It looks like that might be what he tried to do, although obviously without success.
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: Daniel Dravot on April 29, 2013, 02:50:33 AM
To this non-pilot engineer, this seems like an application crying out for modern control technology. Use an IMU (inertial measurement unit) plus GPS in a closed-loop control system to set the throttle, collective and rotor tilt so as to provide whatever state (orientation, position and velocity) the pilot wants as indicated by pair of joysticks. Or an autopilot.

No doubt the purists would consider this "cheating", but I point out that our favorite helicopter-like flying machine, the Apollo LM, worked in much this way. Only the specific mechanisms for producing the reactive forces on the vehicle were different.

I understand that not all states are immediately accessible, e.g., you cannot start translating from a hover without changing your attitude at least momentarily so the pilot would still have to take this into account.

Don't those small hobby helicopters and other flying machines already have systems like this? If so, why not the big machines too?

As a pilot whose air time is better measured in minutes than in hours, and someone who was an engineer a long time ago, I've wondered about this.  Once outside of Sydney, I watched the boat I was on slide more or less sideways to the dock, which seemed impressive to me.  So I wondered, have these control systems become "intelligent," where rather than telling each individual control surface what to do, the captain tells the boat what to do, and the boat figures out how to do it?  Or was it just a highly skilled captain at work?

When Airbus started to produce airplanes which would adjust the control surfaces without asking the pilot's permission first, that seemed to be controversial.  I don't know how it is now.
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: smartcooky on April 29, 2013, 06:07:43 AM
Picture this.

To this non-pilot engineer, this seems like an application crying out for modern control technology. Use an IMU (inertial measurement unit) plus GPS in a closed-loop control system to set the throttle, collective and rotor tilt so as to provide whatever state (orientation, position and velocity) the pilot wants as indicated by pair of joysticks. Or an autopilot.

No doubt the purists would consider this "cheating", but I point out that our favorite helicopter-like flying machine, the Apollo LM, worked in much this way. Only the specific mechanisms for producing the reactive forces on the vehicle were different.

I understand that not all states are immediately accessible, e.g., you cannot start translating from a hover without changing your attitude at least momentarily so the pilot would still have to take this into account.

Other things you have to consider with a helicopter are outside environmental influences. For example, I don't think there is any likelihood of a sudden gust of wind as you are landing your LM in the Fra Mauro highlands.

Also, there is a nasty little phenomenon called "ground resonance". There is no automatic way to resolve this issue, just pilot know-how

Ground resonance will demolish a helicopter very effectively if not dealt with. Here is a short video with an example and explanation.

Quote
Don't those small hobby helicopters and other flying machines already have systems like this? If so, why not the big machines too?

Most of those hobbyist RC helicopters get around the main rotor torque and reaction issues in two ways.

1. they dispense with the swashplate assembly (http://en.wikipedia.org/wiki/Swashplate_%28helicopter%29) that allows main rotor blade pitch angle change and overall rotor tilt control to be combined into a single hub. Instead the main rotor blades are fixed pitch and the rotor only tilts. Lift is controlled using the throttle to vary the main rotor rpm.

2. The torque problem is overcome using two stacked contra-rotating main rotors.

(http://image.wirelessmadness.com/cache/data/RC-Images/Helis/BLH2100UK2-Blade-CX4-RTF-Counter-Rotating-RC-Helicopter-with-2-4Ghz-Transmitter-01-320x320.jpg)

EDIT: Some helicopter types have something called "force trim", which basically allows the pilot to set a new "zero" position for the cyclic control, so for example, he's lifted off and pushed the cyclic forward and is flying forwards at 100 kias. If he were to let go of the cyclic, it would normally come back to the neutral position, however, if he presses the "force trim" button, a new "zero" is set in whatever position the cyclic is in, and if he takes his hand off, it will remain there.
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: Daniel Dravot on April 29, 2013, 06:42:20 AM
Other things you have to consider with a helicopter are outside environmental influences. For example, I don't think there is any likelihood of a sudden gust of wind as you are landing your LM in the Fra Mauro highlands.

Not outside the craft!

Also, there is a nasty little phenomenon called "ground resonance". There is no automatic way to resolve this issue, just pilot know-how

Ground resonance will demolish a helicopter very effectively if not dealt with. Here is a short video with an example and explanation.

Ooh!  The washing machine will do that too, if it is loaded incorrectly.

As a person with precisely zero rotary-wing experience (and very, very slightly more fixed-wing experience), I am surprised they say the correct action is to get back into the air.  I would have thought that when you're on the ground, the corrective action for most bad things that could happen would be, shut it off, now.  Would that not have solved the problem illustrated in the video?
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: Allan F on April 29, 2013, 06:51:39 AM
I see  it as the landing gear bumping on the ground, causes the rotor blades to flex unevenly, which causes a blade to stall, which causes the landing gear to bump on the ground which causes . . . . .

Either get away from the ground, or push the collective down in order to cancel the lift, so the resonnance effect stops. Am I correct?
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: Noldi400 on April 29, 2013, 10:45:26 AM
I had a good friend killed - on TV, no less - when the helicopter he was in crashed while trying to hold a stable hover during a body recovery (they were over a partially demolished water tower; the tail rotor contacted part of the structure and the helo did a 180o pitch-down and hit the ground rotor-first.

Most of my own experience is with military pilots assisting with rescues in rough terrain where the best way to get the patient out is by winch, and they all very much prefer not to try to hold a hover any longer than necessary.  Even without wind, it's just hard to keep everything nulled out and I suspect the same is true in a LM. You remember even Armstrong felt he found himself with some backward and left translational movement just before touchdown and "arrested this backward rate with some possibly spasmodic control motions, but I was unable to stop the left translational rate".  I guess if it was easy, anyone could do it.

Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: Daniel Dravot on April 29, 2013, 11:17:59 AM
I had a good friend killed - on TV, no less - when the helicopter he was in crashed while trying to hold a stable hover during a body recovery (they were over a partially demolished water tower; the tail rotor contacted part of the structure and the helo did a 180o pitch-down and hit the ground rotor-first.

:(

Short version.  From the moon, the Earth is generally within 5o of directly overhead at 0o lat and long. Apollo 17 landed at 20.2o North, 30.8o East, which should put Earth at about 20o from local vertical.

But, all the references from the flight say Earth was at about 45o elevation.  What am I missing that would put the Earth that low in the sky?

Assuming I'm doing this right, the landing spot is almost 29 degrees away from 0-0.  But even if the earth's "wobble" was 5 degrees the other way, that still only gets us to about 34 degrees.

Was the 45 degree figure an actual measurement, or just someone eyeballing it?
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: ka9q on April 29, 2013, 11:44:12 AM
Don't forget that the A17 site was ringed by mountains that put the horizon well above the horizontal in some directions. That's why the earth seems to be at such a low elevation angle in some of their photos; you don't have your inner ear telling you which way is down.
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: Noldi400 on April 29, 2013, 03:26:00 PM
Short version.  From the moon, the Earth is generally within 5o of directly overhead at 0o lat and long. Apollo 17 landed at 20.2o North, 30.8o East, which should put Earth at about 20o from local vertical.

But, all the references from the flight say Earth was at about 45o elevation.  What am I missing that would put the Earth that low in the sky?

Assuming I'm doing this right, the landing spot is almost 29 degrees away from 0-0.  But even if the earth's "wobble" was 5 degrees the other way, that still only gets us to about 34 degrees.

Was the 45 degree figure an actual measurement, or just someone eyeballing it?

From the AS-17 Surface Operations Manual:

(http://i627.photobucket.com/albums/tt353/jarvisn/A17-EarthAzimuth_zpsdb48b9c2.jpg)

And comments in the ALSJ Transcript, talking about orienting the LM high-gain:

115:21:22 Parker: Okay. And, Challenger, that should be a Pitch of 21 and a Yaw of minus 45.

[Cernan - "The high gain (antenna) had a pitch axis and yaw axis. I don't think it had a roll axis. I don't think they were particularly related to the inertial axes of the spacecraft."]
[Gene's memory is correct. From the LM, Earth is at an elevation of 45 degrees and at a bearing 30 degrees south of west; therefore, these pitch and yaw numbers are not simply related to the direction of Earth in the local sky.]

Actually, I think I figured it out by using Orbiter, of all things. 20o N latitude moved Earth's position down about (duh) 20o at 0o longitude; then moving to 30o E longitude moved it down into the right position.

I do wish that my spatial perception skills were better; it seems like I have to draw myself a picture to get the right perception every time.
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: Allan F on September 18, 2013, 07:23:36 PM
On a related note to my original question:

How much did the nozzle of the descent engine heat up during use? I understand the underside and the legs of the descent stage were protected by the heat reflecting kapton and that the radar antenna used to measure distance and closure speed relative to the surface also was protected. The nozzle must have been quite hot. Or did the ablative cooling inside the nozzle prevent the outside of the nozzle to heat up?

I have no idea how to calculate this. The answer would obviously be different relative to the throttle setting and runtime of the engine - and how much of the ablation layer had been burned off. There has to be an equilibrium curve between engine use and temperature, right?

Edit: Also the different parts of the nozzle would have different temperature, as the exhaust cools as it expands.
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: ka9q on September 19, 2013, 07:23:54 AM
Short answer: a lot.

Quite a few in-flight videos of hypergolic engines similar to the LM's show the nozzle glowing orange. The area near the throat glows the brightest, because as you said the exhaust cools as it expands.

A shield was added between the landing radar antenna and descent engine to protect the antenna from the nozzle's thermal radiation. I've also seen it claimed that it was to block a false return from the engine, but I tend to doubt that given how the radars worked - they were both of the CW type, so they already see a lot of their own transmitted signal.
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: Allan F on September 27, 2013, 03:35:08 PM
The LM's descent and ascent engines (and the SPS) were pressure-fed, with supercritical helium supplying the pressure. I understand this system was limited in it's longivity, because heat seeped into the helium tank, raising it's pressure. And when the burst disk blew, the engine was dead.

I Wonder how a similar fueled engine could be kept viable for longer duration missions (Mars). How to put pressure in the tanks without the restraint of keeping helium from heating up and eventually have to be vented. Big plastic bags inside the tanks supplying mechanical pressure?

Cryo-fuels are fine when you have a big industial complex to supply fresh fuel, but in Space, that's not an option - I think. Can a gas be kept cool enough for a long time? Actively cooled?
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: raven on September 27, 2013, 09:22:33 PM
You might need to resort to pumps. If you are going for active cooling, why not go for a full cryogenic propellent setup? Higher ISP (even higher if you go nuclear) that way, which  might help alleviate the the mass of the pumps. You still have to deal with the higher complexity factor though.
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: cjameshuff on September 27, 2013, 10:15:32 PM
I Wonder how a similar fueled engine could be kept viable for longer duration missions (Mars). How to put pressure in the tanks without the restraint of keeping helium from heating up and eventually have to be vented. Big plastic bags inside the tanks supplying mechanical pressure?

A common solution is a gas generator, which could be a pyrotechnic device or something more complex decomposing a liquid fuel.

Cryo-fuels are fine when you have a big industial complex to supply fresh fuel, but in Space, that's not an option - I think. Can a gas be kept cool enough for a long time? Actively cooled?

Cryogenic propellants certainly are an option in space. LOX and CH4 would be relatively easy to keep liquid in the vicinity of Earth or Mars orbit, even with passive cooling. (The WISE instruments reached equilibrium at about 30 K after the cryogen ran out...it took nearly a year for the block of frozen hydrogen it used to sublime away.)
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: ka9q on September 28, 2013, 02:08:08 AM
The LM descent engine had this shelf lifetime problem only because they decided to store the pressurizing helium as a supercritical fluid to reduce mass. You can easily make such a system last indefinitely by just keeping the helium at ambient temperature, and that's exactly what's done on spacecraft that need to produce large delta-Vs after long periods of time, e.g., interplanetary spacecraft arriving at another planet.

The bigger problem is that the important hypergolic propellants freeze at fairly high temperatures. The main problem is N2O4, which freezes at -11.2 C. Straight hydrazine (N2H4) freezes at an even higher temperature, +2 C, which is a challenge for spacecraft that use it as a monopropellant. Bipropellant engines usually use a lower-freezing hydrazine like UDMH (-57 C) or MMH (-52 C), either pure or mixed with straight hydrazine (e.g., Aerozine-50, a 50-50 mix of UDMH and straight hydrazine). I don't know much about its low temperature properties but I've heard it will separate out and cause a problem.
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: Allan F on September 28, 2013, 06:33:05 AM
I see. So the choice to pressurize with supercritical helium was because this system was more compact and lightweight than ambient-temperature storage?

The temperature problem is of course important, but can be managed with solar energy - heat exchangers - maybe even using the tanks as heatsinks for electronics. There is plenty of sunlight (in the inner parts of the solar system) to supply electrical energy.

( I hope you all don't mind me asking all these questions. I have absolutlely no real world use of them, but I enjoy talking to people with skill and information, and I - even at 44 (tomorrow 45) years of age enjoy to get my curiosity satisfied. Even when each answer leads me to ask even more questions )
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: ka9q on September 28, 2013, 10:08:49 AM
That's right, I don't know the details but I did read the supercritical He storage was strictly a mass issue. I guess the lower pressure meant less tank weight. There is also an ambient He tank in the DPS that gets things started but most of it comes from the supercritical tank.
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: VQ on September 29, 2013, 02:48:19 PM
The LM's descent and ascent engines (and the SPS) were pressure-fed, with supercritical helium supplying the pressure. I understand this system was limited in it's longivity, because heat seeped into the helium tank, raising it's pressure. And when the burst disk blew, the engine was dead.

I Wonder how a similar fueled engine could be kept viable for longer duration missions (Mars). How to put pressure in the tanks without the restraint of keeping helium from heating up and eventually have to be vented. Big plastic bags inside the tanks supplying mechanical pressure?

Cryo-fuels are fine when you have a big industial complex to supply fresh fuel, but in Space, that's not an option - I think. Can a gas be kept cool enough for a long time? Actively cooled?
Probably the same way long-duration satellite missions keep their thrusters pressurized. Come to think of it, I am pretty ignorant of how these systems stay pressurized for very long periods of time. No need to use a bladder though - one option would just be to use pressurized helium gas instead of supercritical. Much less space and weight efficient, though.
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: Allan F on September 29, 2013, 06:06:47 PM
Do you need ullage with pressurized tanks? That's not necessary with a bladder system. Helium is such an obnoxius gas - it creeps through the tiniest cracks, and you risk losing your pressure. It is the smallest "particle" of gas, since it is a singular atom, not like hydrogen or oxygen which bond together two by two. If you try to inflate a tire with it, it bleeds right through.
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: ka9q on September 29, 2013, 08:53:27 PM
As I understand it, while He is a very "slippery" gas the main reasons for using bladders is to keep it from dissolving into the propellants and to avoid having it get into the propellant lines in zero-G.
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: Noldi400 on September 29, 2013, 09:32:21 PM
As I understand it, while He is a very "slippery" gas the main reasons for using bladders is to keep it from dissolving into the propellants and to avoid having it get into the propellant lines in zero-G.
Just as a side comment, when looking at how things were done on the LM, it's good to keep in mind that (as you know, Bob...) the bug was very much purpose built - every design feature was oriented to the single purpose of getting two men to the surface and back to orbit (with some exploration & such in between) with as high a probability of success as possible.  The ascent stage, in particular, was designed with as few things to go wrong as possible.  Armstrong had even proposed putting manual valves in the propellant lines so that if the circuits failed, they could fire the engine "by hand".  He says in First Man that "management didn't think that was up to NASA's standards of sophistication".  ::)
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: ka9q on September 30, 2013, 01:30:33 AM
Manual propellant valves seem simple enough, but there are more than just two of them. Besides the valves that actually fire the engine, the tanks are pressurized with helium through one-way valves to keep the propellants from mixing in the manifold. The APS is cross-connected with the RCS (two separate systems) to allow propellant sharing.

Then you have all the other things that have to happen within a split second of ascent, such as firing the guillotine cutters and bolts between the stages.
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: Zach on October 06, 2013, 10:59:37 AM
Manual propellant valves seem simple enough, but there are more than just two of them. Besides the valves that actually fire the engine, the tanks are pressurized with helium through one-way valves to keep the propellants from mixing in the manifold. The APS is cross-connected with the RCS (two separate systems) to allow propellant sharing.

Then you have all the other things that have to happen within a split second of ascent, such as firing the guillotine cutters and bolts between the stages.

It would have been feasible as a last-chance option. A "manual valve" doesn't have to literally open or close a single line, it could be a series of valve functions incorporated into a device that operated with one manual motion. And as far as I'm aware, there isn't any show-stopper that would have prevented firing the guillotines and stage separation devices a few seconds early. The AS would remain in a stable position on the DS until the AS engine fired. Not sophisticated, to be sure, but elegant in its simplicity, especially if it's the difference between being entombed 1/4 million miles from home or being able to see your family again. The decision to not include such an option probably came down to probabilities. If enough stuff had failed to where you're down to that as an option, what is the likelihood that manually firing the engine would result in a successful rescue?
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: ka9q on October 09, 2013, 12:37:00 AM
I always wondered if the ascent stage really would remain stable without the attach bolts, cables and piping. Some of the LMs landed at decidedly non-level attitudes.

Whether it makes sense to implement something like manual APS valve control really depends on an understanding of the various failure mechanisms and probabilities. Our intuition about such things is often wrong, and it doesn't make sense to spend effort (and more importantly, mass and volume) on backups for things unlikely to fail when other fatal failures are much more likely.
Title: Re: Is my math correct? LM descent engine force at nozzle
Post by: Noldi400 on October 12, 2013, 06:56:11 PM
I always wondered if the ascent stage really would remain stable without the attach bolts, cables and piping. Some of the LMs landed at decidedly non-level attitudes.

Whether it makes sense to implement something like manual APS valve control really depends on an understanding of the various failure mechanisms and probabilities. Our intuition about such things is often wrong, and it doesn't make sense to spend effort (and more importantly, mass and volume) on backups for things unlikely to fail when other fatal failures are much more likely.
True enough, but the idea belonged to Armstrong, who should know those things as well or better than anyone. The statement may have been influenced by the fact that is was made around 2003 (to James R. Hansen), and with Neil's dry wit he may have been halfway joking. The actual passage reads:

Quote
"We did have various means of controlling the circuitry to the valves - opening the flow of propellants to the engine. So that was an alternative. I had proposed many months earlier - maybe even years earlier - that we just put a big manual valve in there to open those propellant valves rather than, or in addition to, having all the electronic circuitry. But management didn't think that was up to NASA's standards of sophistication.  So I really knew that circuitry very well. But it wasn't really a problem, because if we fired the engine and it didn't fire, we weren't out of time. We had a lot of time to think about the problem to figure out what else we could do. When pilots get really worried is when they run out of options and out of time simultaneously.*"
-FIRST MAN, James R. Hansen
* Like, say, trying to miss a boulder field and a giant crater while your fuel runs low and your computer acts up?