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Post by moonglow on Jun 22, 2005 2:19:52 GMT -4
Kiwi, thank's for the welcome. Well the copy I noticed it on I taped a long time ago on the Discovery channel and has pretty good contrast. On my For all Mankind DVD it's more washed out and I can hardly make out the texture of the mountain behind the LM, so I guess it's possible the copy you have you can't see the texture vary well. But if what I'm seeing is dust on the len's that would make since of what I saw. The black sky is what I meant looking like a mask because it looked like it moved down over the mountain's texture as the camera zoomed out, but again if that's dust on the len's instead of texture that would explain it. I did consider that what I was seeing was TV snow or something but it looked more like texture to me but I'm willing to concede that it's not.
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Bob B.
Bob the Excel Guru?
Posts: 3,072
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Post by Bob B. on Jun 22, 2005 10:24:41 GMT -4
Thanks L.O. and B.B.. i would be very interested in the calcs. I don't have any calculations on the hammer drop thing, but I can show you how I arrived at the fall times for the particles. First, let's determine the fall time of a particle in a vacuum. This is an easy calculation, which I'm sure you can do for yourself, but I'll show anyway. Let's say the particle is dropped from a height of 2 meters, we have: on Earth, t = (2d/a)^0.5 = (2*2/9.807)^0.5 = 0.64 s on the Moon, t = (2*2/1.62)^0.5 = 1.57 s However, because of drag, soil-sized particles may actually fall slower on Earth, which is exactly the opposite of what the HBs claims. A small particle will reach its terminal velocity very quickly (in only a fraction of a second), thus we can assume the particle will fall at a constant velocity (the terminal velocity) over the entire distance. From this we can determine the fall time. The lunar soils are mainly classified as sandy-silt or silty-sand. Therefore, let's assume a particle with a diameter of 10 microns, which falls into the size range of silt. If we assume a density of 2700 kg/m^3, which is fairly typical of rocky material, we can calculate the mass of the particle, r = 10 / 1E+6 / 2 = 0.000005 m v = 4/3*pi*r^3 = 4/3*pi*0.000005^3 = 5.236E-16 m^3 m = v*density = 5.236E-16*2700 = 1.414E-12 kg The cross-sectional area of the particle is, A = pi*r^2 = pi*0.000005^2 = 7.854E-11 m^2 The drag force on a particle is given by, Fd = Cd*rho*A*V^2/2 where Cd is the drag coefficient, rho is the air density, A is the area normal to the flow, and V is the velocity. We know that F=ma, thus we can write, ma = Cd*rho*A*V^2/2 Terminal velocity occurs when the acceleration caused by the drag force is equal to the acceleration of gravity. In this condition we have a net acceleration of zero and the velocity remains constant. We therefore replace the acceleration, a, in the above equation with the acceleration of gravity, g, giving us, mg = Cd*rho*A*V^2/2 Rearranging variables we get, V = [ 2mg/(Cd*rho*A) ]^0.5 which is the terminal velocity. We know that g equals 9.807 m/s^2, and the standard density of air at sea level is 1.225 kg/m^3. In my previous calculations I used a value of 0.5 for Cd, which is the normal coefficient of friction for a sphere. However, this number is generally higher for very small particles. Furthermore, a soil particle is not a perfect sphere. Nonetheless, let's use Cd=0.5 knowing that our calculation is conservative, i.e. the time we calculate will be the *fastest* time in which the particle will fall. We have, V = [ 2*1.414E-12*9.807 / (0.5*1.225*7.854E-11) ] ^0.5 = 0.759 m/s Thus the fall time is, t = d/v = 2/0.759 = 2.64 s I previously calculated 2.69 s because I took into consideration the time needed to accelerate to terminal velocity. This level of precision isn't needed in your case, thus there is no need to over-complicate things. Don't forget that if we use a higher value of Cd, the fall time increases. For instance, if Cd=1 then fall time becomes 3.73 seconds. I remember when I posted this information originally, Dr. John (jrkeller) found a chart showing drag coefficient versus particle size, but I don't recall the link. And on more thing, these calculations assume the air is completely still. In reality, very small particles will usually remain in suspension longer because of air currents. Now that you have an example, you can vary the particle size, the soil density, and the drag coefficient to calculate other examples. I think these computations show quite convincingly (at least to a non-HB) that small soil particles will fall more slowly on Earth than they will on the Moon.
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Post by jaydeehess on Jun 22, 2005 17:43:27 GMT -4
Beautiful Bob, I will be cutting and pasting much of this. No worries, I can follow the math, it's like being in undergrad physics again, but my typing skills are not great I have thought of another addendum, to this. If one drives down a dirt road when it is dry you need not be going very fast at all in order to kick up a great cloud of dust behind the vehicle. This dust cloud remains in the air for quite some time due to it's reaching it's terminal velocity , which as a result is also very small, quickly,,, AND because the turbulent air following the vehicle keeps lifting the dust. IF the lunar vehicle was on Earth , in an atmosphere the dust would behave exactly the same way. The lack of a following dust cloud proves that this was not on Earth and in an atmosphere.
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Post by jaydeehess on Jun 22, 2005 18:10:45 GMT -4
Here is how I put it on the other forum;
If an object is allowed to fall on Earth it accellerates in obeyance to Newton's formula a=F/m. But this will happen only until the force due to air resistance equals that of gravity. The drag force is (Cd*rho*V^2 )/2 Cd= coefficient of drag rho=density of air V=velocity
So when ma=(Cd*rho*V^2)/2 the onject will be at terminal velocity.
We can see that heavier objects will have a larger terminal velocity that tiny ones and indeed we know that a dust particle or a feather falling horizontally will reach it's , slower , terminal velocity faster than say a hammer. (BTW drop the feather with the point of the qill aimed straight down and for a short distance(several feet) both the feather and hammer will drop pretty close to the same rate).
In an atmosphere dust drops slower than in no atmosphere obviously since a dust particle has very little mass and therefore reaches it's terminal velocity very quickly.
First, let's determine the fall time of a particle in a vacuum. This is an easy calculation, which I'm sure you can do for yourself, but I'll show anyway. Let's say the particle is dropped from a height of 2 meters, we have:
on Earth, t = (2d/a)^0.5 = (2*2/9.807)^0.5 = 0.64 s
on the Moon, t = (2*2/1.62)^0.5 = 1.57 s
However, because of drag, soil-sized particles may actually fall slower on Earth, which is exactly the opposite of what the HBs claims. A small particle will reach its terminal velocity very quickly (in only a fraction of a second), thus we can assume the particle will fall at a constant velocity (the terminal velocity) over the entire distance. From this we can determine the fall time.
The lunar soils are mainly classified as sandy-silt or silty-sand. Therefore, let's assume a particle with a diameter of 10 microns, which falls into the size range of silt. If we assume a density of 2700 kg/m^3, which is fairly typical of rocky material, we can calculate the mass of the particle,
r = 10 / 1E+6 / 2 = 0.000005 m
v = 4/3*pi*r^3 = 4/3*pi*0.000005^3 = 5.236E-16 m^3
m = v*density = 5.236E-16*2700 = 1.414E-12 kg
The cross-sectional area of the particle is,
A = pi*r^2 = pi*0.000005^2 = 7.854E-11 m^2
The drag force on a particle is given by,
Fd = Cd*rho*A*V^2/2
where Cd is the drag coefficient, rho is the air density, A is the area normal to the flow, and V is the velocity. We know that F=ma, thus we can write,
ma = Cd*rho*A*V^2/2
Terminal velocity occurs when the acceleration caused by the drag force is equal to the acceleration of gravity. In this condition we have a net acceleration of zero and the velocity remains constant. We therefore replace the acceleration, a, in the above equation with the acceleration of gravity, g, giving us,
mg = Cd*rho*A*V^2/2
Rearranging variables we get,
V = [ 2mg/(Cd*rho*A) ]^0.5
which is the terminal velocity.
We know that g equals 9.807 m/s^2, and the standard density of air at sea level is 1.225 kg/m^3. In my previous calculations I used a value of 0.5 for Cd, which is the normal coefficient of friction for a sphere. However, this number is generally higher for very small particles. Furthermore, a soil particle is not a perfect sphere. Nonetheless, let's use Cd=0.5 knowing that our calculation is conservative, i.e. the time we calculate will be the *fastest* time in which the particle will fall. We have,
V = [ 2*1.414E-12*9.807 / (0.5*1.225*7.854E-11) ] ^0.5 = 0.759 m/s
Thus the fall time is,
t = d/v = 2/0.759 = 2.64 s
Don't forget that if we use a higher value of Cd, the fall time increases. For instance, if Cd=1 then fall time becomes 3.73 seconds.
Now that you have an example, you can vary the particle size, the soil density, and the drag coefficient to calculate other examples. I think these computations show quite convincingly that small soil particles will fall more slowly on Earth than they will on the Moon.
One more thing, these calculations assume the air is completely still. In reality, very small particles will usually remain in suspension longer because of air currents.
Thus, if one drives down a dirt road when it is dry you need not be going very fast at all in order to kick up a great cloud of dust behind the vehicle. This dust cloud remains in the air for quite some time due to it's reaching it's terminal velocity , which as a result is also very small, quickly,,, AND because the turbulent air following the vehicle keeps lifting the dust.
IF the lunar vehicle was on Earth , in an atmosphere, the dust would behave exactly the same way. The lack of a following dust cloud proves that this was not on Earth and in an atmosphere.
In contrast to this an astronaut jumping into the air on Earth would fall about 2.5 times faster than that same astronauts jumping up to the same height on the surface of the Moon.
Now, would one of you rocket scientists please find me an example of either moon dust falling too slow (ie. no atmosphere) or an astronaut falling to fast. (ie. Earth gravity) note to AP readers, the 'rocket scientist' part I found personally ironic considering my source of info
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Post by jaydeehess on Jun 22, 2005 18:19:07 GMT -4
The absolute most rediculous HB claim ever just has to be Una Ronald's Coke bottle on the Moon.
One woman tells a story and there is absolutly NO corroborating evidence that anything she has stated is in fact true. She has her times wrong, she has referred to letters to the editor that do not exist and NO ONE else in the entire continent of Australia, or the world for that matter, saw this "Coke bottle" that was, according to her immediatly recognizable and very clear!!!
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Bob B.
Bob the Excel Guru?
Posts: 3,072
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Post by Bob B. on Jun 22, 2005 19:05:49 GMT -4
Beautiful Bob, I will be cutting and pasting much of this. I hope it helps you win your argument. ;D Here's something else to think about. Soil is made up of particles of many different sizes, each with a different terminal velocity. If you throw a hand full of dirt in the air, the particles will segregate with the largest falling to the ground very quickly and the smallest lingering in the air for several seconds before settling. You don't see this at all in the Apollo footage. All the soil particles fall at the same rate without any segregation.
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Post by jaydeehess on Jun 22, 2005 23:06:13 GMT -4
Good point, and this may even be the cause of the HB complaint that it,"doesn't look right".
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Post by Obviousman on Jun 23, 2005 7:40:03 GMT -4
I always like that 'coke bottle' claim.
I, along with thousands of other schoolkids, watched the moonwalk live DURING THE DAY in Perth.
(The egress started just before 1000 local, 20 JUL 69, in Perth. We watched a live feed, not a 'replay')
There was NO coke bottle.
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Post by jaydeehess on Jun 24, 2005 19:03:26 GMT -4
I always like that 'coke bottle' claim. I, along with thousands of other schoolkids, watched the moonwalk live DURING THE DAY in Perth. (The egress started just before 1000 local, 20 JUL 69, in Perth. We watched a live feed, not a 'replay') There was NO coke bottle. IIRC Bennett and Percy 'adjust' Una's story since they know that what she saw was not live. They assume that she watched a replay late at night(though there is no proof of such a replay) and that a 'whistleblower' introduced the bottle in the tape of the replay which afterwards was removed by NASA. Una claims she was watching it live and never has changed that. Bennett and Percy say she thought she was watching it live and had every expectation to believe it was live. (though the latter requires that she was living in a cave prior to the event IMHO)
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Post by PhantomWolf on Jun 24, 2005 21:47:31 GMT -4
Which takes it to a really surreal place. If we take the idea that she saw a repeat and the coke bottle had been added, then the obvious "bottle adder" would have been a comedian at the TV Station airing it, not a "NASA whistleblower" nd any lter showing would be the original simply because the Station Mnager would hve fired the comedian and had the original restored, all without NASA even being involved. But no, overlook the obvious because NASA has to be involved with everything. Oh look my cat sneeze, it must be an areol pollutate sprayed by NASA, run for your lives......
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Post by Kiwi on Jun 27, 2005 6:13:00 GMT -4
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Post by Kiwi on Jun 27, 2005 6:27:29 GMT -4
Naturally it includes appropriate praise for members of this board: quote: -------------------------------------------------------------------------------- Originally posted by dknjms: ...it's actually stuff from a subsequent landing... that has lead me to wonder) but for some reason no-one ever wants to talk about those. -------------------------------------------------------------------------------- (Emphasis added.) You obviously haven't been to the BABB or Apollohoax! At those places you'll find plenty of people (JayUtah and Bob B. in particular) who are so passionate about Apollo that the hardest job you'd ever have there is getting them to shut up... <Fixed typos>
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Post by blackpudding on Jun 30, 2005 16:38:46 GMT -4
with regard to the tire not being able to be filled with air.
There is no reason why tires or balloons etc can't be filled with air/nitrogen/CO2 etc(from canisters). All that is important is the difference in pressures between the inside of the tire and the outside.
At sea level the pressure difference is the difference between the pressure on the inside of a car tire and atmospheric pressure(out side the tire).
On the moon there would be no atmospheric pressure so all you have to do is fill the tore(or inner tube) with the amount of gas(what ever they used) to create a pressure difference between the inside of the tore and nought(the pressure on the outside).
The tire would only explode if you filled it up too much, otherwise it would behave normally.
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Post by echnaton on Jun 30, 2005 17:07:45 GMT -4
with regard to the tire not being able to be filled with air. There is no reason why tires or balloons etc can't be filled with air/nitrogen/CO2 etc(from canisters). All that is important is the difference in pressures between the inside of the tire and the outside. At sea level the pressure difference is the difference between the pressure on the inside of a car tire and atmospheric pressure(out side the tire). On the moon there would be no atmospheric pressure so all you have to do is fill the tore(or inner tube) with the amount of gas(what ever they used) to create a pressure difference between the inside of the tore and nought(the pressure on the outside). The tire would only explode if you filled it up too much, otherwise it would behave normally. Welcome to the board, blackpudding Some of the less knowledgeable hoax proponents seem of be of the opinion that a vacuum creates a sort of large or even infinite energy gradient. Thus space suits must balloon up or tires will just explode or whatever. Since they are really quite ignorant it is hard to determine exactly what will cause this or why they think it happens.
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Post by JayUtah on Jun 30, 2005 18:48:11 GMT -4
It's all about the restraint layers. Hydraulic equipment such as skip loaders employ flexible hoses carrying pressures of several hundred pounds per square inch. Yet they don't balloon up. Why? Because there is impregnated in them a metal mesh that accepts the inflation load and provides the tube with appropriate tensile strength. The rubber or plastic component simply provides impermeability.
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