Fun with shadows - again! The Taurus-Littrow fly-by

[bknight]

That's the same formula as my eqn. 2 and there is no specific angle at which the approximation "reverts" to the exact formula. The error is a function of the angle which is always nonzero, except for the trivial case of theta = 0. That's why I provided the one percent error example.

Yes I read that percent error comment.  Now take the following table using the different formula variations
theta   10*57.3/theta   10/tan(theta/57.3) percent difference
0.5   1145.91559   1145.886501   0.002538491
1   572.9577951   572.8996163   0.01015412
2   286.4788976   286.3625328   0.040618956
4   143.2394488   143.0066626   0.162515437
8   71.61972439   71.15369722   0.650696678
16   35.8098622   34.87414444   2.613016917

So where would it be prudent to say the angle is too large for the abbreviated form of the relationship to be valid?
.5%  1%  2%?

[12oh2alarm]

So where would it be prudent to say the angle is too large for the abbreviated form of the relationship to be valid?
.5%  1%  2%?

Depends on what you want to do with the result, i.e. what error is acceptable.
For determining whether we can see the CSM shadow, I'd even allow a 10% error. It's a back-of-the-envelope calculation.
If Goldstone points their antenna, .5% is way too big an error and no doubt they will not use the tan x = x approximation in anything they compute.

[12oh2alarm]

Seemed to me that, in the presence of bright sunshine, there might just be a shadow of a command module to be seen somewhere.
[...]
What I need to know is are the numbers right and are my conclusions reasonable?

For anyone who decides to check the 16mm footage, most of the dark blobs that appear to trail the movement of the CSM & LM seem more likely to what we in the trade call 'bits of crap'.

The laws of perspective require that if you draw lines from points to their shadows, they must meet in a single point (which may be off the image). Try to do that with some of the mountaintops. Now draw a line from the CSM to the suspected shadow. They don't match.

This can be used to more easily identify shadows, since you know you must look only on the line from the CSM to the common intersection of all shadow lines.
But as calculated in another post, don't get your hopes up. The shadow of a CSM is only visible on the moon shortly before lithobreaking... 
If you want to burn some time, I'd compare old lunar surface images to those from LRO and see if a new crater has appeared in the last 40 years. If you're the first to spot one, it might get your name attached. How's that for vanity?
Monkey Crater, yeah!

[bknight]

Depends on what you want to do with the result, i.e. what error is acceptable.
For determining whether we can see the CSM shadow, I'd even allow a 10% error. It's a back-of-the-envelope calculation.
If Goldstone points their antenna, .5% is way too big an error and no doubt they will not use the tan x = x approximation in anything they compute.

Fair enough and that is probably the best answer.

[ka9q]

If you want to burn some time, I'd compare old lunar surface images to those from LRO and see if a new crater has appeared in the last 40 years. If you're the first to spot one, it might get your name attached. How's that for vanity?

If you're going to do this, use the LOIRP (Lunar Orbiter Image Recovery Project) datasets for your 1960s references. They're digitized directly from the wideband tapes that recorded the raw downlink signal. It's basically what we could have done with the Apollo 11 tapes if they hadn't been recycled in the early 1980s.

[bknight]

If you're going to do this, use the LOIRP (Lunar Orbiter Image Recovery Project) datasets for your 1960s references. They're digitized directly from the wideband tapes that recorded the raw downlink signal. It's basically what we could have done with the Apollo 11 tapes if they hadn't been recycled in the early 1980s.

Well NASA had t erase the tapes to avoid "fakery discovery" 

[ChrLz]

The laws of perspective require that if you draw lines from points to their shadows, they must meet in a single point (which may be off the image). Try to do that with some of the mountaintops. Now draw a line from the CSM to the suspected shadow. They don't match.

While that's true in theory, camera lenses generally do not capture an accurate 'linear' image..  So, especially when the field of view is wide, this method may fall down - as the entire scene is curved..  The camera is effectively a point in space and the scene around it is a sphere.  That issue is why Xavier Pascal, Jack White and others get it so horribly wrong when they point out that shadows and sun-flares don't necessarily point at the sun, especially in stitched panoramas.  Such scenes involve very wide angles and that's where you simply cannot use straight lines...

[smartcooky]

The laws of perspective require that if you draw lines from points to their shadows, they must meet in a single point (which may be off the image). Try to do that with some of the mountaintops. Now draw a line from the CSM to the suspected shadow. They don't match.

While that's true in theory, camera lenses generally do not capture an accurate 'linear' image..  So, especially when the field of view is wide, this method may fall down - as the entire scene is curved..  The camera is effectively a point in space and the scene around it is a sphere.  That issue is why Xavier Pascal, Jack White and others get it so horribly wrong when they point out that shadows and sun-flares don't necessarily point at the sun, especially in stitched panoramas.  Such scenes involve very wide angles and that's where you simply cannot use straight lines...

THIS IN SPADES!!!

This is why its obvious to me that HBs who bang on about non-parallel shadows do not have the foggiest understanding of photography.

There is one simple thing to keep in mind that helps people to understand this...

"a photograph is a two-dimensional representation of a three-dimensional reality"

[raven]

If you want to burn some time, I'd compare old lunar surface images to those from LRO and see if a new crater has appeared in the last 40 years. If you're the first to spot one, it might get your name attached. How's that for vanity?

If you're going to do this, use the LOIRP (Lunar Orbiter Image Recovery Project) datasets for your 1960s references. They're digitized directly from the wideband tapes that recorded the raw downlink signal. It's basically what we could have done with the Apollo 11 tapes if they hadn't been recycled in the early 1980s.

I really wish that hadn't needed to be done. I've seen Polaroids and 8mm film taken of the pre-kinescoped video, and there's really quite a difference in the contrast and detail.

[bknight]

...

There is one simple thing to keep in mind that helps people to understand this...

"a photograph is a two-dimensional representation of a three-dimensional reality"

Exactly.  I was "discussing" some of the Earth shots from A8 with a YT'er and (s)he indicated that the photos were bogus because the continents didn't look like the Gall–Peters projection.  I tired to explain a map is a 2 dimension object of a roughly spherical 3 dimension object.  The continents wouldn't look that way when viewed from space, to bad he didn't get it.

[12oh2alarm]

Exactly.  I was "discussing" some of the Earth shots from A8 with a YT'er and (s)he indicated that the photos were bogus because the continents didn't look like the Gall–Peters projection.  I tried to explain a map is a 2 dimension object of a roughly spherical 3 dimension object.  The continents wouldn't look that way when viewed from space, to bad he didn't get it.

Must be too hard for an HB to find a globe, view it from the same direction, and compare the result. You could even do it keeping the scale correct, i.e. from N earth radii away. Doing basic research is not their strong point it would seem.

[12oh2alarm]

Time for some visual math (aka geometry).

The deepest penumbra shadow is cast when the CSM is directly overhead. What would it look like looking up from the ground? The CSM's circular lunar orbit is 110km above ground.
It appears under an angle of 2 * atan(5m/110.000m) = 0.0052 degrees, which is almost exactly 1/100 of the Sun's angular diameter.
Lets picture that:


The dip in the light curve on the ground is 1 part in 10.000 or less. Compare this to a Venus transit:



or Mercury transit:

[bknight]

Must be too hard for an HB to find a globe, view it from the same direction, and compare the result. You could even do it keeping the scale correct, i.e. from N earth radii away. Doing basic research is not their strong point it would seem.

You expect HB's to do basic research???

[12oh2alarm]

You expect HB's to do basic research???

I'm old. I'm gentle. I'm patient. Everybody is assumed clever until proven stupid. Everybody gets a second chance.

[bknight]

You expect HB's to do basic research???

I'm old. I'm gentle. I'm patient. Everybody is assumed clever until proven stupid. Everybody gets a second chance.

This person was on his fourth or fifth try.  Then there was the "you can't see stars in the daytime because the atmosphere was opaque, due to sunlight being scattered(?) not sure of the reasoning.