Joined: May 2005 Gender: Male Posts: 3,072 Location: Dayton, Ohio
Re: The "DPS should have dug a blast crater" Meme « Reply #30 on Jun 30, 2011, 11:46pm »
I'm bumping this old thread to let you know that I finally got around to writing up a formal analysis of the blast crater issue. If you're interested, you can read it at the link below. I hope this is adequate to put the issue to rest.
I'm bumping this old thread to let you know that I finally got around to writing up a formal analysis of the blast crater issue. If you're interested, you can read it at the link below. I hope this is adequate to put the issue to rest.
Thanks for that Bob. I do like your very succint conclusion:
Despite the limitations of this analysis, the conclusion is inescapable: no pronounced crater is formed. No more soil can be removed than there is energy available to detach, entrain, and transport it away. The energy of the exhaust gas has been determined to a reasonably high degree of confidence, and the energy require to produce a large crater is simply not present.
The part in bold is what I have been trying to explain to a certain individual.
Joined: May 2005 Gender: Male Posts: 3,122 Location: Space City, Texas
Re: The "DPS should have dug a blast crater" Meme « Reply #33 on Jul 1, 2011, 3:09pm »
Great work, Bob. Your comprehensive web site has grown to the point that it is time to make a separate discussion section on the forum, like the section for discussions of Jay's Clavius site.
Re: The "DPS should have dug a blast crater" Meme « Reply #34 on Jul 1, 2011, 7:50pm »
Good work!
In a couple of places you compute the exhaust gas energy, e.g., 16.36 MJ just before landing. That appears to be the energy in one second's worth of exhaust gas, so shouldn't that instead be given as the power in the exhaust gas, e.g., 16.36 megawatts (MW)?
In a couple of places you compute the exhaust gas energy, e.g., 16.36 MJ just before landing. That appears to be the energy in one second's worth of exhaust gas, so shouldn't that instead be given as the power in the exhaust gas, e.g., 16.36 megawatts (MW)?
I think you're correct. I'm going to have to go back through the page and figure out what corrections need to be made.
EDIT
OK, I think I got that part fixed but I just noticed another even bigger mistake. Just before the section titled "Mapping the Exhaust Stream" I perform the following calculation
Not only only should that be power I'm calculating, but I also need to multiply the added energy (from the expansion of the gas) by the mass flow rate. The added energy was in J/kg and the mass flow rate in kg/s, therefore I have J/kg × kg/s = J/s, or watts. The calculation should be,
The worse part of it is that I carried this mistake through all my subsequent calculations. It doesn't change the results dramatically - the maximum depth changes from 36.7 mm to 39.9 mm - but I have to redo my illustrations. Those things were a pain in the neck to produce and now I've got to do it all over. Crap!
Joined: May 2005 Gender: Male Posts: 3,072 Location: Dayton, Ohio
Re: The "DPS should have dug a blast crater" Meme « Reply #37 on Jul 2, 2011, 7:22pm »
OK, I think I've got it all fixed up. Thanks ka9q for pointing out my mistake. I had lost sight of the fact that I was multiply by a mass flow rate rather than just mass. My second mistake would have probably gone unidentified had you not forced me to go back and double check all my calculations.
In Figure 4 the units are correctly in joules because I'm multiplying the power of the exhaust stream by the duration of exposure.
One of these days I'll figure out the difference between pressure thrust and momentum thrust...
It’s really pretty simple. The best way to explain it is with an example. Let’s say we have an engine with the following characteristics:
Propellant flow rate: 100 lbm/s Exhaust gas velocity: 10,000 ft/s Diameter of nozzle exit: 3.5 ft Exhaust pressure at nozzle exit: 1 psi Ambient air pressure: 0 (vacuum)
Most of the thrust comes from the momentum of the exhaust gas, which is the mass flow rate times the exhaust gas velocity,
Fmomentum = (100 / 32.174) × 10,000 = 31,081 lbf
In addition, unbalanced pressure forces at the nozzle exit produce a small amount of thrust. This is calculated by
Area of nozzle exit × (exhaust pressure at nozzle exit – ambient air pressure)
In our case, the area of the nozzle exit is 1,385 in2, so we have
Fpressure = 1,385 × (1 – 0) = 1,385 lbf
Our total thrust is,
Ftotal = 31,081 + 1,385 = 32,466 lbf
The engine specific impulse is,
Isp = 32,466 / 100 = 324.66 s (vacuum)
In may problems it’s common to use the effective exhaust gas velocity, which we get by multiplying the specific impulse by standard gravity,
C = 324.66 × 32.174 = 10,446 ft/s
As you can see, this isn’t the true exhaust gas velocity because pressure thrust is lumped in with it. When working with Tsiolkovsky's rocket equation we want to use C; however, in my blast crater analysis I needed to use the real exhaust velocity because that’s what determines the kinetic energy of the gas.
Joined: Jul 2006 Gender: Male Posts: 582 Location: Atlanta, GA
Re: The "DPS should have dug a blast crater" Meme « Reply #39 on Nov 29, 2011, 11:47am »
Since this issue has popped up again, I'll take this opportunity to add my thanks to Bob B. for putting another nail in the coffin of the Apollo Hoax theory.
Re: The "DPS should have dug a blast crater" Meme « Reply #40 on Nov 29, 2011, 7:03pm »
Thanks Bob, I meant to acknowledge this sooner.
Maybe it would help if I can think about where the thrust is being applied. I assume that most of the thrust is transferred to the top of the combustion chamber due to the imbalance in pressure between the top and bottom of the chamber. But what about the nozzle? Does it transfer the pressure thrust to the vehicle?
Brilliant. Why didn't someone do this earlier. I am so sick of the blast crater nonsense.
Just one thing, could you reference the source for the bulk density of the lunar soil for completeness sake.
Fine work sir.
Yeah. Hoax Theorists erroneously work under the impression that the LM was being fired against the lunar surface at full power in an atmosphere where gas doesn't rapidly expand away.
Re: The "DPS should have dug a blast crater" Meme « Reply #42 on Feb 16, 2012, 5:24pm »
Question concerning reply #9, these calculations consider the effect of the descent engine exhaust to be uniform over an area, over a specific time. The scope of this type of calculation must be limited only to the force within an area where the exhaust stream is unhindered. What is missing in these calculations is the inclusion of the factor of compression of the exhaust gas caused by the continuous flow of exhaust gas being forced to move through a confined area, with respect to time.
After the exhaust gas initially contacts the lunar surface, a cm3 area of gas at the center will than move outward, say 1 cm, as it does a cm3 of gas must be combined into this cm3 area as a result of the continued flow of exhaust gas, this additional cm3 of gas must be compressed into the original cm3 area and so on until the original cm3 of gas reaches the perimeter of the plume, 137 cm from the center. The original cm3 of exhaust gas will now have accumulated 138 cm3 of exhaust gas compressed into a 1 cm3 area.
Can the compression of the exhaust gas be formally addressed and calculated?
After the exhaust gas initially contacts the lunar surface, a cm3 area of gas at the center will than move outward, say 1 cm, as it does a cm3 of gas must be combined into this cm3 area as a result of the continued flow of exhaust gas, this additional cm3 of gas must be compressed into the original cm3 area and so on until the original cm3 of gas reaches the perimeter of the plume, 137 cm from the center. The original cm3 of exhaust gas will now have accumulated 138 cm3 of exhaust gas compressed into a 1 cm3 area.
It doesn't work that way. Besides, it's irrelevant to calculating the kinetic energy of the gas. Bottom line...
Kinetic energy of gas < Kinetic energy needed to produce a crater
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