Cross-bedding on Mars?

"Daryl Krupa" wrote in message
om...
" George" wrote in message
t...
"Daryl Krupa" wrote in message
om...
" George" wrote in message
. ..
"Daryl Krupa" wrote in message
om...
" George" wrote in message
.. .
snip

NASA spent some 800 million dollars on the project. I am
confortable
believing that they know the characteristics of the optics or
the
cameras.
snip

Ah, yes, but do _you _know?

Actually, I do.

Yet, you have not given any indication that you do,
other than your simple assertion of superior knowledge.

Gee guy, it has nothing to do with whether I am smarter or dumber than
anyone else. It has to do with understanding how the pictures are
created,
and how to recreate them.

No, George, to determine scale in and of an image,
you have to know some important details about the geometry of the system.
You have not demonstrated such knowledge, so
your assertions image scale are contestable.

No ****? I didn't know I was taking a final exam in optics 101. Try going
to JPL's web site and getting the dimensions of the rover. You really don't
know anything about optics or geometry, do you?

http://athena.cornell.edu/pdf/tb_pancam.pdf

http://www.space.com/scienceastronom...3D_040210.html

snip
In fact, I know that they took these issues into account
when they designed and tested the craft.

Do you indeed.
You cannot use claims of superior knowledge on the part of others
to support your claim that you yourself have superior knowledge.
IOW, you have not demonstrated that should be regarded as
an authority, i.e a reliable source of information, on this subject.
It would be much more useful to simply quote something that
a genuine authority has written on this subject.

Daryl Krupa

Ok. Don't take my word for it. Go to the following links and read
about
it:

http://www.highmars.org/niac/education/mer/mer00.html

http://www.atsnn.com/marscolors.html

Not a whit at either of those sites image scale.

I wasn't posting about scale. I was posting about how to create the images.
Try going to JPLs web site and looking up the dimensions of the rover. Have
you ever made a scale map from a photograph? If you know the optics of the
camera, and you know the scale of the image, and the exact size of an object
in the image, you can create a 2-d scale map approximation of the image. Of
course it will have variations, and inaccuracies with an inexpensicve setup.
I have done this many times, and then verified the results ini the field,
and found it to be pretty accurate for most purposes. Do you honestly think
that NASA send two rovers worth 850 million dollars to Mars without knowing
how they were going to determine scale in the images? How do you think they
figured out the height of the bedrock at the Opportunity site?

http://athena.cornell.edu/pdf/tb_pancam.pdf

http://www.space.com/scienceastronom...3D_040210.html

Another way you can determine the approximate size of an object in an image
is by knowing the image size, and the distance to the object. If the image
size is 14.2225 inches on each side (a square, obviously), and the object in
the image is 5.2 cm in distance from the camera, and the apparent size of
the object in the image is 6 inches based on the measurements of the object
in the image, you tell me what the closest approximate object size is. Can
you do the math?

By the way, I've published some low-resolution copies (96 dpi) of some
of
the RGB color images I created using the pancam filtered image files, if
you
are interested. I have much higher resolution images, but not enough
web
space topublish them. If you are interested in obtaining the higher
resolution images (up to 600 dpi), let me know how to get them to you,
and
we can discuss that.

I am not interested in your processing of images.
I am interested in how you are able to derive image scale for those
images.

snip diversionary tactic


Fine. You try to be nice to people, and all they want to do is show how
much of an ass they can be. I provided with some links above that should
point you in the direction that you want to go. If that is not good enough
for you, I suggest that you contact Cornell University, since they helped
design the damned thing:

http://athena.cornell.edu/reference/contact.html

If that is not good enough for you, I suggest you go have some sex, because
you are being about as anal as a person can be, and appear to be deficient
in this area.

snip
(Kenneth Chiu) wrote in message ...
It seems like there is sufficient information. Have I
overlooked something?

Yes. For instance, an indication of how one might determine image
scale.

snip

The big missing piece is range, but you can get that for
relatively near objects from the stereo.

Indeed one can, if one has the stereo pair for comparison.
How is that done with a single iamge that is not displayed
with the other half of its stereo pair?
And what about distant objects?
And what about the microsopic images?
Are they in stereo pairs?

Whoops, missed this.

The two PanCam eyes are separated by 30 cm horizontally and
have a 1? toe-in. This separation and toe-in provide an
adequate convergence distance for scientifically useful
stereo topographic and ranging solutions to be obtained
from the near-field (5-10 m) to approximately 100 m from
the rover.

So, since we know how many radians there are per pixel, and
we know the distance, we can get the size for anything from
10m to 100m. Does that seem right?

It might; what do you mean by "the distance"?
If you mean the distance between the "eyes", then yes,
I suppose it might be possible, but it would be nice to have a formula
available for doing that.
If you mean the distance between an eye and an object in an image,
how do we know that?

Daryl Krupa

" George" wrote in message . ..
I wasn't posting about scale.
snip

Oh, yes, you were.
It is to this that I responded:

BEGIN QUOTE

From: " George"
Newsgroups: sci.geo.geology,sci.astro

Subject: Cross-bedding on Mars?

Message-ID:
X-Trace: ldjgbllpbapjglppdbdpiflmbcekedmfhojhikkbagflhcbodb loicodgebgmbfffhafmoopdadpoljhfbkoppggbfibgnfiihan ackflekmlilolnmbhlcnknekmbegkhjmnbjcjipammiobhfoki pbhgdhggfd
NNTP-Posting-Date: Sat, 21 Feb 2004 22:35:16 EST
Date: Sat, 21 Feb 2004 22:40:05 -0500

"Ralph Nesbitt" wrote in message
m...

" George" wrote in message
. ..


http://marsrovers.jpl.nasa.gov/galle...P2932M1M1.HTML

I took several long hard looks at this one the day it was first posted
on
the JPL website, and several times afterwards. First of all, the
resolution
is not high enough in the image to make a solid determination was to
what
it
represents. I even increased the resolution to 400 dpi, increased the
contrast, and the image intensity, and the best I can determine what you
have there is either gas pockets or pitting, as in all the other damaged
specimens. They are definitely not chambers. Secondly, there are many
other photos of these spheres, and none of them have shown any internal
biologic structure. Nearly all of them have the same solid internal
granular crystalline structure that we see on the outside.

Unless I missed it, there is no size scale posted for pic's from the "Mars
Rovers". A scale for size perspective with these Pic's would be great
IMHO.
Ralph Nesbitt


Actually, if you look at enough of the images, you will note that
parts of
the rover is present some of them. This does allow one to calibrate
one's
perspective of scale for many of the others. It may not be precise,
but it
works in a pinch.

END QUOTE

So how do you do that scale calculation, George?

Daryl Krupa

In article ,
Daryl Krupa wrote:
snip
(Kenneth Chiu) wrote in message ...
It seems like there is sufficient information. Have I
overlooked something?

Yes. For instance, an indication of how one might determine image
scale.

Sorry, I think I've lost the original point of discussion in
this thread. So you are looking for an explanation of how
to calculate size from the geometric/optical parameters? Or
only evidence that NASA has the necessary information?


snip

The big missing piece is range, but you can get that for
relatively near objects from the stereo.

Indeed one can, if one has the stereo pair for comparison.
How is that done with a single iamge that is not displayed
with the other half of its stereo pair?

Ah, well, you would need a stereo pair obviously. You could
get that from the NASA web site.

And what about distant objects?

It seems that more than 100 meters would be a problem.
Probably not enough separation.

And what about the microsopic images?

Those are not stereo, but have very shallow depth-of-field,
so if it's in focus, you know the range. The spacing per
pixel is available on the Web and was posted previously to
the newsgroups.

Are they in stereo pairs?

Whoops, missed this.

The two PanCam eyes are separated by 30 cm horizontally and
have a 1? toe-in. This separation and toe-in provide an
adequate convergence distance for scientifically useful
stereo topographic and ranging solutions to be obtained
from the near-field (5-10 m) to approximately 100 m from
the rover.

So, since we know how many radians there are per pixel, and
we know the distance, we can get the size for anything from
10m to 100m. Does that seem right?

It might; what do you mean by "the distance"?

Distance to the object.

If you mean the distance between the "eyes", then yes,

I suppose it might be possible, but it would be nice to have a formula
available for doing that.
If you mean the distance between an eye and an object in an image,
how do we know that?

Okay, I'll give it a shot.

For small angles the width of the object is just a*r, where
a is angular displacement and r is the distance to the
object.

We know that the pancam is 0.28 mrad/pixel, from a previous
link. (I'm going to assume that mrad stands for
milliradians, though I'm not positive about that). So if
something is one pixel wide, and 10 meters away, that means
it's about 3 millimeters wide.

To get the range to a point, we need the stereo pairs. The
off-axis angle of a line to the point is arcsin(d/f), where
d is the displacement of the point in the image plane, and f
is the focal length.

So we take one line from each image in the pair, and from
the toe-in and interocular separation, we calculate the
intersection of the two lines to get the range.

If you use variables for everything, you should be able to
get a formula out of it. (Or a least an equation that you
can solve numerically.)

It's been a decade since I've done calculations like this,
so you should double-check this before actually trying it.
(Maybe someone else can verify it.)

"Daryl Krupa" wrote in message
om...
snip
(Kenneth Chiu) wrote in message
...
It seems like there is sufficient information. Have I
overlooked something?

Yes. For instance, an indication of how one might determine image
scale.

snip

The big missing piece is range, but you can get that for
relatively near objects from the stereo.

Indeed one can, if one has the stereo pair for comparison.
How is that done with a single iamge that is not displayed
with the other half of its stereo pair?

Daryl. If you had bothered to go to the mission web site, you will note
that they are posting both images of the stereo pairs.

And what about distant objects?

The range for the rovers is 100 meters. That is the published range.
Anything farther can be measured with the orbiting satellites.

And what about the microsopic images?
Are they in stereo pairs?

Yes, many of them are.

Whoops, missed this.

The two PanCam eyes are separated by 30 cm horizontally and
have a 1? toe-in. This separation and toe-in provide an
adequate convergence distance for scientifically useful
stereo topographic and ranging solutions to be obtained
from the near-field (5-10 m) to approximately 100 m from
the rover.

So, since we know how many radians there are per pixel, and
we know the distance, we can get the size for anything from
10m to 100m. Does that seem right?

It might; what do you mean by "the distance"?
If you mean the distance between the "eyes", then yes,
I suppose it might be possible, but it would be nice to have a formula
available for doing that.
If you mean the distance between an eye and an object in an image,
how do we know that?

Daryl Krupa

Take a class in analytical geometry, dude.

"Kenneth Chiu" wrote in message
...
In article ,
Daryl Krupa wrote:
snip
(Kenneth Chiu) wrote in message
...
It seems like there is sufficient information. Have I
overlooked something?

Yes. For instance, an indication of how one might determine image
scale.

Sorry, I think I've lost the original point of discussion in
this thread. So you are looking for an explanation of how
to calculate size from the geometric/optical parameters? Or
only evidence that NASA has the necessary information?


snip

The big missing piece is range, but you can get that for
relatively near objects from the stereo.

Indeed one can, if one has the stereo pair for comparison.
How is that done with a single iamge that is not displayed
with the other half of its stereo pair?

Ah, well, you would need a stereo pair obviously. You could
get that from the NASA web site.

And what about distant objects?

It seems that more than 100 meters would be a problem.
Probably not enough separation.

And what about the microsopic images?

Those are not stereo, but have very shallow depth-of-field,
so if it's in focus, you know the range. The spacing per
pixel is available on the Web and was posted previously to
the newsgroups.

I misstated this earlier. I said that the micros were stereo, but they are
not. My mistake. Sorry guys.

Are they in stereo pairs?

Whoops, missed this.

The two PanCam eyes are separated by 30 cm horizontally and
have a 1? toe-in. This separation and toe-in provide an
adequate convergence distance for scientifically useful
stereo topographic and ranging solutions to be obtained
from the near-field (5-10 m) to approximately 100 m from
the rover.

So, since we know how many radians there are per pixel, and
we know the distance, we can get the size for anything from
10m to 100m. Does that seem right?

It might; what do you mean by "the distance"?

Distance to the object.

If you mean the distance between the "eyes", then yes,

I suppose it might be possible, but it would be nice to have a formula
available for doing that.
If you mean the distance between an eye and an object in an image,
how do we know that?

Okay, I'll give it a shot.

For small angles the width of the object is just a*r, where
a is angular displacement and r is the distance to the
object.

We know that the pancam is 0.28 mrad/pixel, from a previous
link. (I'm going to assume that mrad stands for
milliradians, though I'm not positive about that). So if
something is one pixel wide, and 10 meters away, that means
it's about 3 millimeters wide.

To get the range to a point, we need the stereo pairs. The
off-axis angle of a line to the point is arcsin(d/f), where
d is the displacement of the point in the image plane, and f
is the focal length.

So we take one line from each image in the pair, and from
the toe-in and interocular separation, we calculate the
intersection of the two lines to get the range.

If you use variables for everything, you should be able to
get a formula out of it. (Or a least an equation that you
can solve numerically.)

It's been a decade since I've done calculations like this,
so you should double-check this before actually trying it.
(Maybe someone else can verify it.)

"Daryl Krupa" wrote in message
om...
" George" wrote in message
. ..
I wasn't posting about scale.
snip

Oh, yes, you were.
It is to this that I responded:

BEGIN QUOTE

From: " George"
Newsgroups: sci.geo.geology,sci.astro

Subject: Cross-bedding on Mars?

Message-ID:
X-Trace:
ldjgbllpbapjglppdbdpiflmbcekedmfhojhikkbagflhcbodb loicodgebgmbfffhafmoopdadp
oljhfbkoppggbfibgnfiihanackflekmlilolnmbhlcnknekmb egkhjmnbjcjipammiobhfokipb
hgdhggfd
NNTP-Posting-Date: Sat, 21 Feb 2004 22:35:16 EST
Date: Sat, 21 Feb 2004 22:40:05 -0500

"Ralph Nesbitt" wrote in message
m...

" George" wrote in message
. ..



http://marsrovers.jpl.nasa.gov/galle...P2932M1M1.HTML

I took several long hard looks at this one the day it was first posted
on
the JPL website, and several times afterwards. First of all, the
resolution
is not high enough in the image to make a solid determination was to
what
it
represents. I even increased the resolution to 400 dpi, increased the
contrast, and the image intensity, and the best I can determine what
you
have there is either gas pockets or pitting, as in all the other
damaged
specimens. They are definitely not chambers. Secondly, there are
many
other photos of these spheres, and none of them have shown any
internal
biologic structure. Nearly all of them have the same solid internal
granular crystalline structure that we see on the outside.

Unless I missed it, there is no size scale posted for pic's from the
"Mars
Rovers". A scale for size perspective with these Pic's would be great
IMHO.
Ralph Nesbitt


Actually, if you look at enough of the images, you will note that
parts of
the rover is present some of them. This does allow one to calibrate
one's
perspective of scale for many of the others. It may not be precise,
but it
works in a pinch.

END QUOTE

So how do you do that scale calculation, George?

Daryl Krupa

Simple answer? NASA has stated that the "blueberries" are about 2-3 mms.
If I assume they are correct (And I do, since if I was being paid to do
this, I would be making the calculations - I'm not, so I don't, and I'm also
not being paid to double-check their numbers). If you assume that NASAs
estimate of 2-3 mms is correct (and I do make that assumption) I simply
measure the "blueberry" on the image, and determine the size of the 'pits'
based on the measurement of the object (making the assumption that 2 mm in
real life is proportional to what ever the image size of the object is that
I am measuring on the picture, and I scale it accordingly. No calculus is
needed. Just simple arithmatic. While this 'fudge factor" is obviously not
precise by NASAs or by most engineering standards, it works in a pinch for
estimating purposes. One thing you can be certain of, if one assumes that
the "blueberry'" is about 2.5 mm, and the pit is less or equal to 1/4 of the
size of the object, then it is certainly less than or equal to 0.625 mm,
give or take the fudge factor.

In article ,
George wrote:

"Kenneth Chiu" wrote in message
...
In article ,
Daryl Krupa wrote:
snip
(Kenneth Chiu) wrote in message
...
It seems like there is sufficient information. Have I
overlooked something?

Yes. For instance, an indication of how one might determine image
scale.

Sorry, I think I've lost the original point of discussion in
this thread. So you are looking for an explanation of how
to calculate size from the geometric/optical parameters? Or
only evidence that NASA has the necessary information?


snip

The big missing piece is range, but you can get that for
relatively near objects from the stereo.

Indeed one can, if one has the stereo pair for comparison.
How is that done with a single iamge that is not displayed
with the other half of its stereo pair?

Ah, well, you would need a stereo pair obviously. You could
get that from the NASA web site.

And what about distant objects?

It seems that more than 100 meters would be a problem.
Probably not enough separation.

And what about the microsopic images?

Those are not stereo, but have very shallow depth-of-field,
so if it's in focus, you know the range. The spacing per
pixel is available on the Web and was posted previously to
the newsgroups.

I misstated this earlier. I said that the micros were stereo, but they are
not. My mistake. Sorry guys.

Hmm...actually you are right. It is not a stereo imager,
but can be used to get stereo images. Near the end of:

http://athena.cornell.edu/pdf/tb_micro.pdf

"Kenneth Chiu" wrote in message
...
In article ,
George wrote:

"Kenneth Chiu" wrote in message
...
In article ,
Daryl Krupa wrote:
snip
(Kenneth Chiu) wrote in message
...
It seems like there is sufficient information. Have I
overlooked something?

Yes. For instance, an indication of how one might determine image
scale.

Sorry, I think I've lost the original point of discussion in
this thread. So you are looking for an explanation of how
to calculate size from the geometric/optical parameters? Or
only evidence that NASA has the necessary information?


snip

The big missing piece is range, but you can get that for
relatively near objects from the stereo.

Indeed one can, if one has the stereo pair for comparison.
How is that done with a single iamge that is not displayed
with the other half of its stereo pair?

Ah, well, you would need a stereo pair obviously. You could
get that from the NASA web site.

And what about distant objects?

It seems that more than 100 meters would be a problem.
Probably not enough separation.

And what about the microsopic images?

Those are not stereo, but have very shallow depth-of-field,
so if it's in focus, you know the range. The spacing per
pixel is available on the Web and was posted previously to
the newsgroups.

I misstated this earlier. I said that the micros were stereo, but they
are
not. My mistake. Sorry guys.

Hmm...actually you are right. It is not a stereo imager,
but can be used to get stereo images. Near the end of:

http://athena.cornell.edu/pdf/tb_micro.pdf

The reason I changed my mind was because I had tried to produce a 3-D image
using the images from the microscopic imager, and the ones I produced didn't
work very well. But I let my wife look at them (she is also a geologist)
and she thought that they looked ok. I just don't have very good eyesight,
so 3-D images are hard for me.