Pointless Test - Just For Fun

Quote: Face

+93C or -129C. Buffalo's practically southern Canada, so I knew this one ;)

Oh, so you know first hand how lunar night feels like. :)

Quote:

Did the astronauts that were actually there have issues

Not as far as heat or cold are concerned, that I know of. All missions took place in lunar day time, BTW. The issue most mentioned was lunar dust. They tracked a great deal of it into the LEMs, and it was messy. Very fine, very abrasive.

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I'd think the space suit was fully self contained, or they at least gained/lost heat at a painfully slow rate.

Something like that. Sunlight will only warm the outermost layer of the suit anyway. How much of that gets through depends on what kind f insulation you have. As I recall, though, the Apollo suits incorporated a liquid coolant system.

Quote: Nareed

Oh, so you know first hand how lunar night feels like. :)

Woke up today to an actual (not wind chill) temp of -9F/-23C. Call it lunar dusk ;)

Quote: Nareed

In one word: no.

On Earth, the atmosphere scatters sunlight, so there is some even in the shade. You might have enough to power something using solar panels, but not very well. On the airless Moon, however, shadow is about the same as night. Not exactly, because some light is reflected off the surface, creating a glare all over the sunlit side. That's nowhere near enough to provide power to, say, a solar-powered calculator. It would be like trying to power it with starlight, or for that matter with moonlight on Earth.

This is debatable. You are correct that on earth, atmosphere scatters sunlight. We can than agree that the atmosphere on the moon, is almost non-existent. So there is almost no scattering of light. Right?

Now, the earth, with its lush atmosphere, oceans, and clouds, reflects a lot more light from the sun, to space and other objects (in this example the moon). It reflects more light than the moon does to Earth. Correct? This is non-debatable.

Now, on the moon, regardless of location (assuming fairly flat natural landscape, no cliffs, craters, caves) during the night phase, an ample amount of light is received from the reflection of light from the sun off the Earth. This should be enough to power most solar based tools.

This is info I read months back on an article about astronomy and solar activity. I just compacted it from memory. There are some other details that I can't remember about light transmission, but solar powered tools will work, on the moon, in the night phase, BETTER than solar powered tools, on earth, during our night phase.

It is that reversal of the atmospheres that decreases light scatter, and increases reflection.

So, in one word: yes.

Quote: YoDiceRoll11

This is info I read months back on an article about astronomy and solar activity. I just compacted it from memory. There are some other details that I can't remember about light transmission, but solar powered tools will work, on the moon, in the night phase, BETTER than solar powered tools, on earth, during our night phase.

Assuming this is so, and I'm not willing to just grant it is so, you forget another thing. While the Earth is always on most of the Moon's day sky, it's not always full. It shows phases as much as the Moon does. Maybe solar powered gear would work under a full Earth's light. But what of a half Earth or a crescent or waxing Earth?

Oh, and every so often you'd get a "New Earth" reflecting no light at all.

So: it depends. That's two words if you're keeping count ;)

Quote: Nareed

Assuming this is so, and I'm not willing to just grant it is so, you forget another thing. While the Earth is always on most of the Moon's day sky, it's not always full. It shows phases as much as the Moon does. Maybe solar powered gear would work under a full Earth's light. But what of a half Earth or a crescent or waxing Earth?

Oh, and every so often you'd get a "New Earth" reflecting no light at all.

So: it depends. That's two words if you're keeping count ;)

You are correct!! Good observation. Of course I was making assumptions in line with the question at the beginning of this thread. :)

So in a word: Maybe.

:P

Quote: YoDiceRoll11

You are correct!! Good observation.

Thanks.

As we see the sky from Earth, it's easy to forget other worlds have a sky of their own. For example, from Mars you'd see the Earth and Moon as separate "stars" some of the time (likewise from Venus, if you could peer through the clouds; I'm not sure about Mercury).

Quote: Nareed

Thanks.

As we see the sky from Earth, it's easy to forget other worlds have a sky of their own. For example, from Mars you'd see the Earth and Moon as separate "stars" some of the time (likewise from Venus, if you could peer through the clouds; I'm not sure about Mercury).

Maybe one day, they will open up a casino on Mars. Drinks on me.

@

The Red Rock!

There is no information on whether you are facing the earth. The same part of the moon is always facing the earth, that is, on the moon, there is no earth rise and no earth set. So, either the earth is there or it is not. You don't have enough information.

However, the average albedo of the earth is 0.3, which means 70% of its radiation received is beamed back into space. At that point, if you were to consider the earth a radiation source, how much "solar" radiation would you receiving, presuming the earth was full.

Well, the earth receives about 1,360 Watts/m^2 and reradiates 70%, say, 950 watts/m^2. The moon is about 384,000,000 meters from the earth. From this, you can calculate the received from the earth on the moon.

We can use the

I = E(4*Pi*R^2)/(4*Pi*r^2)

I = Irradiance at the surface of the outer sphere
E = Irradiance at the surface of the object (Sun)
4 x R2 = surface area of the object
4 x r2 = surface area of the outer sphere

I = 950*(4*Pi*6378000^2)/(4*Pi*384000000^2) = .262 W/m^2 (presuming that the earth is directly overhead). So, the light from a reflected earth would receive something of the order of 1,200 times less radiation than direct sunlight (on the earth)

Consider a 100W light bulb. How far away from a light bulb would you need to be to receive the same radiation?

.262 = 100/(4*Pi*r^2)
r = sqrt (30.319) = 5.50 meters.

Try powering your solar calculator with a light bumb being 20 feet away from a 100 watt light bulb. So I would say anything solar powered on the dark side would be fairly useless, but you could at least see.

The next question is regarding the gun. How much force does shooting a .45 gun give you. E = 1/2 mv^2. At 15 grams and 270 m/s speed, the kinetic energy is 270^2 * 1/2 * .015 = 562J. That translates into a kinetic energy in the opposite direction, except your mass is 100kg (energy is preserved). 562J = 1/2mv^2.

v^2 = 562/50 = 112m/s. So, theoretically, that bullet would have you moving at 10.6 m/s or about 38km/hr. This is not insignficant. So, you jump in the air (you can jump up about 10 feet), fire the gun, and see how far you get before landing. At 10 feet up, it would take you about 3 seconds to fall (force of gravity is 1.62 m / 2^2), you firing a bullet could take you about 32 meters or about 100 feet. Or you could take a run, fire the gun and see how far you get. If you could get your speed up to 5 m^2, take a jump and fire the bullet, maybe you could get yourself 150 feet. Of course you would need about 7,000 bullets to get you all 200 miles!

And finally, the FM radio only works line of sight. You would know the frequency of the mother ship. I would assume that the mother ship would erect an antennae out of materials to be able to listen for things. The distance you can see is sqrt(h(2R+h). So, for someone 2 meters high on a moon with a radius of 1737.4km, the FM radio would have an effective broadcast of (2*(1737400+2)^.5 = 2.636km. A 20 meter tower has a range of 26.362km. A 50 meter tower haw a range of about 66km.

Quote: boymimbo

The next question is regarding the gun. How much force does shooting a .45 gun give you. E = 1/2 mv^2. At 15 grams and 270 m/s speed, the kinetic energy is 270^2 * 1/2 * .015 = 562J. That translates into a kinetic energy in the opposite direction, except your mass is 100kg (energy is preserved). 562J = 1/2mv^2.

v^2 = 562/50 = 112m/s. So, theoretically, that bullet would have you moving at 10.6 m/s or about 38km/hr. This is not insignficant. So, you jump in the air (you can jump up about 10 feet), fire the gun, and see how far you get before landing. At 10 feet up, it would take you about 3 seconds to fall (force of gravity is 1.62 m / 2^2), you firing a bullet could take you about 32 meters or about 100 feet.

You almost got it boymimbo. 270 meters a second is NOT enough to propel a 100kg person 32m into the air after a 3m jump (rounded).

The situation you want to consider is much simpler than that. I'm going to work in 1 dimension because we want to track just the vertical ... You're on the right path with energy conservation, but you neglected the potential energy component. The applicable equation here is KE0 + PE0 = KEf + PEf (inital and final potential and kinetic energies, respectively). That's the actual definition of conserved energy. In this case, you want maximum height. That happens the exact moment you stop moving upward, so we are looking for a final kinetic energy of zero. Also, we know that PE = mgh (mass, gravity, height) so we need to solve for PEf to get final height. We're also assuming that our astronaut can jump 10ft up, which I rounded to 3m, making PE0 = (100kg)(1.62m/s^2)(3m) = 486J. Transferred kinetic energy from the bullet is KE0 = (1/2)(.015)(270m/s)^2 = 546.75J, so PEf = 486J + 546.75J = 1032.75J. Invoking the definition of potential energy, Hf = PEf / (mg) = 1032.75 / (1.62 * 100) = 6.34m.

Working backwards assuming a final height of 32m, the bullet would have to leave the gun at about 791.5 m/s to get our hero astronaut to the desired height.

Quote: weaselman

The real problem with this idea is thar gun powder isn't anything like a signal flare - it needs oxygen to burn.

Actually it's a lot like a signal flare, both use anaerobic propellants. If it was aerobic, there just wouldn't be enough oxygen in the cartridge to deliver any meaningful amount of energy.
There are some potential issues guns can have in vacuum, such as vacuum stiction that can disrupt lower-force automatics, but they are relatively minor.

Quote: YoDiceRoll11

You're on the right path with energy conservation, but you neglected the potential energy component. The applicable equation here is KE0 + PE0 = KEf + PEf (inital and final potential and kinetic energies, respectively).

It's not that either.
Energy conservation has nothing to do with it. Mechanical energy is not conserved in most processes, but converted to heat and other forms of energy. What is conserved is impulse.

The impulse from a .45 ACP pistol is calculated simply as m*v and equals 4 N*s. Gas energy should be added, at another 0.4 N*s. Translated to a 100kg body, it would provide a velocity of 0.044m/s.

If you used .45 Win Mag, its impulse is 7 kg*m/s for factory ammunition. My reloading table shows that by using a custom load with stock bullets and a 6" barrel you could get 8 kg*m/s. If you really went out of your way to cast custom flat-nose bullets filled with DU, you could can reach 11.2 N*s from the bullet and 12.8 N*s total.

Completely discharging two conventional 7+1 round M1911 style .45 ACP pistols would net you 35 N*s of impulse or just a bit over 1 foot per second.
If you had two USPSA frame pistols chambered for .45 Win Mag with custom heavy bullet loads and extended double-stacks with 17+1 rounds each, discharging both pistols would net you 460 N*s or 4.6 m/s, and if you had two spare mags for each, 13.3 m/s.

That would be a big deal for propulsion in outer space, assuming you can be sure not to hit your station, but on the Moon gravity won't let you make use of it. With a deceleration of 1.6 m/s, your 4.6 m/s without reloading would be consumed in just 3 seconds, and that's firing two very unpleasantly recoiling guns.

Gunpowder does not need oxygen to burn! didnt you see The Good, The Bad, and the Ugly?

BTW is it not a fact that the "dark side of the moon" is a misnomer? The moon rotates and the "dark" side gets just as much light, more even [due to eclipses].

I didn't say that shooting the gun made us go any higher. I would you the gun to propel you FORWARD, not to jump HIGHER. You take a jump, fire the gun so it propels you in the opposite direction, and land. There is no delta z in my equation, just the height. If you fired the gun standing on the ground, you would go nowhere, as your feel are planted (not so firmly) to the ground. Take the gun, jump up, fire the gun opposite in the direction you want to go, and land.

Quote: boymimbo

I didn't say that shooting the gun made us go any higher.

Quote: boymimbo

So, you jump in the air (you can jump up about 10 feet), fire the gun, and see how far you get before landing. At 10 feet up, it would take you about 3 seconds to fall (force of gravity is 1.62 m / 2^2), you firing a bullet could take you about 32 meters or about 100 feet.

So without even specifying an angle, you calculated 100 feet from......what? If it was lateral, than that makes the math more complicated, but it comes to the same conclusion. You can't possibly make a 100 foot lateral movement from the firing of one round in the opposite direction.

It sounds like you were clearly talking about vertical height.

Please clarify if I missed something because your initial statement was vague than.

Edit: double post. Ignore.

Quote: P90

It's not that either.
Energy conservation has nothing to do with it. Mechanical energy is not conserved in most processes, but converted to heat and other forms of energy. What is conserved is impulse.

The impulse from a .45 ACP pistol is calculated simply as m*v and equals 4 N*s. Gas energy should be added, at another 0.4 N*s. Translated to a 100kg body, it would provide a velocity of 0.044m/s.

If you used .45 Win Mag, its impulse is 7 kg*m/s for factory ammunition. My reloading table shows that by using a custom load with stock bullets and a 6" barrel you could get 8 kg*m/s. If you really went out of your way to cast custom flat-nose bullets filled with DU, you could can reach 11.2 N*s from the bullet and 12.8 N*s total.

Completely discharging two conventional 7+1 round M1911 style .45 ACP pistols would net you 35 N*s of impulse or just a bit over 1 foot per second.
If you had two USPSA frame pistols chambered for .45 Win Mag with custom heavy bullet loads and extended double-stacks with 17+1 rounds each, discharging both pistols would net you 460 N*s or 4.6 m/s, and if you had two spare mags for each, 13.3 m/s.

That would be a big deal for propulsion in outer space, assuming you can be sure not to hit your station, but on the Moon gravity won't let you make use of it. With a deceleration of 1.6 m/s, your 4.6 m/s without reloading would be consumed in just 3 seconds, and that's firing two very unpleasantly recoiling guns.

While you know a great deal about firearms, you make some poor conclusions regarding the physics. Your conclusion is sound but mechanical energy has barely anything to do with this question. "Mechanical Energy" is generally used to refer to the more general concept of "work" as a part of some kind of process whereby energy is being converted from one form to another, like an engine or a reactor, which could be said to produce "mechanical" energy in the form of a rotating output shaft or something similar. While there is mechanical action in a semi-auto, it doesn't really equal mechanical energy output.

Conservation of impulse is not a thing (mass times velocity is momentum). Momentum does have to be conserved, but that's not how the original question was phrased, nor is it the correct way to come to an answer.

No matter, our conclusions are very similar so no worries in the end. You can't make it an extra 100 feet either vertically or laterally. Math doesn't lie.
The above is more of a thought process, not trying to make anything personal. You have a lot more rep on here than I do P90, respect.

Cheers.

Quote: YoDiceRoll11

"Mechanical Energy" is generally used to refer to the more general concept of "work" as a part of some kind of process whereby energy is being converted from one form to another, like an engine or a reactor, which could be said to produce "mechanical" energy in the form of a rotating output shaft or something similar.

I guess I used some terms in an incorrect way, from physics terminology standpoint. TBH I don't remember most of it by now, and there are some substantial differences between physics, engineering and colloquial usage.

Duh, I was conserving energy, not momemtum.

Quote: boymimbo

Duh, I was conserving energy, not momemtum.

And yet you said you could still make up 100 ft of distance.........

No worries guys. No more moon questions. Ha.

I should stick with my dice. Too many physics classes....

Did nobody see Wall-E? Use the rope to strap the oxygen tanks to your suit, get a running start, and then fly there using the tanks as thrusters.



Mythbusters did a test where they demonstrated that an air tank could crash through a cinderblock wall, and while they failed to successfully propel a speedboat (in water, on Earth), they did get it to travel 120 feet at a top speed of 5 knots (roughly 5.75mph). Given that (a) your weight on the moon is a lot less than a speedboat on Earth, and (b) your drag on the moon is virtually non-existent compared to a speedboat in water, I figure you could simply launch yourself toward the mothership and build up enough speed to get to the mothership in a few hours.

Anyone know what the thrust is out of a 100lb tank of compressed oxygen?
Edit: bring the guns for course correction.

Quote: MathExtremist

Did nobody see Wall-E? Use the rope to strap the oxygen tanks to your suit, get a running start, and then fly there using the tanks as thrusters.



Mythbusters did a test where they demonstrated that an air tank could crash through a cinderblock wall, and while they failed to successfully propel a speedboat (in water, on Earth), they did get it to travel 120 feet at a top speed of 5 knots (roughly 5.75mph). Given that (a) your weight on the moon is a lot less than a speedboat on Earth, and (b) your drag on the moon is virtually non-existent compared to a speedboat in water, I figure you could simply launch yourself toward the mothership and build up enough speed to get to the mothership in a few hours.

Anyone know what the thrust is out of a 100lb tank of compressed oxygen?

Not a bad idea. Wouldn't have any drag from the water. The problem would be to remain airborne so not to hit any large craters. But even using it as a sort of hovercraft, even for a short time, would be possible.

Quote: MathExtremist

Did nobody see Wall-E?

I'm not nobody, but I didn't see it.

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Use the rope to strap the oxygen tanks to your suit, get a running start, and then fly there using the tanks as thrusters.

Why bother? Use the implied portable jet pack stored along with the implied spacesuits :P

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Anyone know what the thrust is out of a 100lb tank of compressed oxygen?

I think it depends on the size of the tank. Gasses are highly compressible, after all. A large 100 lbs tank would use lower pressure than a small one, say, giving you different thrust levels.

But I'm not sure of this.

Quote:

Edit: bring the guns for course correction.

Good. Another can of worms :)

Quote: MathExtremist

Given that (a) your weight on the moon is a lot less than a speedboat on Earth, and (b) your drag on the moon is virtually non-existent compared to a speedboat in water, I figure you could simply launch yourself toward the mothership and build up enough speed to get to the mothership in a few hours.

You have gravity drag, and your only way to remain airborne is to constantly combat it - air cushion won't reasonably work due to the vacuum outside.

That could certainly be done, but you'll need dozens of oxygen tanks to cover 200 miles. With just two tanks and no special equipment, you'll cover a couple miles and suffocate.
If you do have special equipment, the energy contained in high-pressure tanks is significant, about 1MJ/kg, or 100MJ in all. It would be OK if you had a gas-powered car, but in flight you'd be shooting it at 1km/s (if you have a special nozzle system that can do that), getting 10,000N*s of momentum per kilogram or 900,000 total.

For your weight of 160N per man, not counting tank and equipment, that's about 1 minute of flight per kilogram per man. And you have 3-4 men, in spacesuits, plus the tanks, for a total of ~600kg and ~1000N. Let's suppose your first 100,000 N*s are spent on acceleration to 100m/s, the next 750,000 on flight, and the remainder on braking (low gravity doesn't reduce impact damage); that's 750s of flight or 75km.
So you can only cover less than 1/4 of the way, even if you have a special nozzle system to use the gas at maximum efficiency, which you don't. Just opening the valve will not send the air out at Mach 3 speed in the desired direction.


By the way, since we're talking about oxygen tanks - where is replacement scrubber for your spacesuits? You can't absorb oxygen all at once, only 5% per breath, and spacesuits contain a rebreather to remove carbon dioxide. It's not a cryogenic rebreather (only one model was ever produced), so CO2 is removed chemically by combining it with LiOH scrubber. You produce over 1kg of CO2 per day, essentially 1.4 times your O2 consumption (C+2O). (6+16+1)g of LiOH can remove (12+16*2)g of CO2, approx. 1:2, so you'd need a full 140 lbs to make full use of your oxygen tanks. That isn't a whole lot, but it's more than stored in your spacesuits. Apollo suits were only good for 7 hours of autonomy.

Quote: P90

That could certainly be done, but you'll need dozens of oxygen tanks to cover 200 miles. With just two tanks and no special equipment, you'll cover a couple miles and suffocate.

I agree.

Quote: Nareed

Good. Another can of worms :)

Actually using the guns for course correction in a low(er) drag environment is more feasible than the previous hypothesis.

Quote: P90

For your weight of 160N per man, not counting tank and equipment, that's about 1 minute of flight per kilogram per man. And you have 3-4 men, in spacesuits, plus the tanks, for a total of ~600kg and ~1000N. Let's suppose your first 100,000 N*s are spent on acceleration to 100m/s, the next 750,000 on flight, and the remainder on braking (low gravity doesn't reduce impact damage); that's 750s of flight or 75km.

I missed the part about there being other people.

It sounds like if I were by myself, the two air tanks might get me there if I:
a) used the first to launch myself upward into the lunar sky toward the mothership, then discarded it
b) "floated" ballistically toward the target, using the second tank to propel me upward and further as needed.
c) just before crash-landing, inflate the life raft upside down and land on it like the jump cushions used by fire departments.

I can't readily do the conversions, but would the energy in an air tank be sufficient to launch it (not you) into orbit? If so, launch an air tank into lunar orbit with the solar powered radio set to repeat a message with your coordinates (which you determined via looking at the star map) and tell the mothership to come get you...

Quote: MathExtremist

I missed the part about there being other people.

It sounds like if I were by myself, the two air tanks might get me there if I:
a) used the first to launch myself upward into the lunar sky toward the mothership, then discarded it
b) "floated" ballistically toward the target, using the second tank to propel me upward and further as needed.
c) just before crash-landing, inflate the life raft upside down and land on it like the jump cushions used by fire departments.

I can't readily do the conversions, but would the energy in an air tank be sufficient to launch it (not you) into orbit? If so, launch an air tank into lunar orbit with the solar powered radio set to repeat a message with your coordinates (which you determined via looking at the star map) and tell the mothership to come get you...

HAHAHA Best answer,........

Ever.

Quote: MathExtremist

It sounds like if I were by myself, the two air tanks might get me there if I:
a) used the first to launch myself upward into the lunar sky toward the mothership, then discarded it
b) "floated" ballistically toward the target, using the second tank to propel me upward and further as needed.

Still far from enough. At least just using the tanks. If you had a special nozzle and control system to go with them, then *maybe*. And tanks would have to be very high-pressure and nearly weightless (when empty), not anything of the modern era.

It's very far from orbital velocity too. Compressed air has only ~1/10 the energy density of fuel+oxidizer. And lunar orbital velocity is 1750m/s, which is still at the limit of ordinary rocket capabilities.

But most importantly you need a nozzle. Without one, you'll waste more than 90% of air's energy and won't get further than a couple miles, if you're lucky. Then you'll crash and die, unless you had a good way to stop.

Quote: Face

The Director of my department is a thinking-type, and wants others in command to be likewise. In interviews for management positions.................

Well, it looks like we have plenty of scientists and engineers on the moon, trying to rectify their plight. However, since the OP was looking at management, I would respond that from Houston control, I am not experiencing any issues, and btw, my plan was flawless had poor execution not messed it up.