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Author Topic: Roger Craig  (Read 101113 times)

Offline Jack Trojan

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Re: Roger Craig
« Reply #352 on: February 13, 2021, 02:09:27 AM »
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Jack....Will you assist me in analyzing this photo?



P.S.....This plea is open to anybody who is willing to honestly debate the in situ photo.

Question # 1....  Can you tell me the dimensions of the box above the rifle in the photo?

I won't draw this out for you, however, you need to measure the total length of 1 of your MCs from the end of the stock to the end of the barrel. The image of the MC is 332 pixels, which = X inches. When I measured the MC minus the barrel I got 300 pixels / 34 inches = 8.8 pixels/inch. This tells me that a rifle with the barrel should be 37.7 inches. Let me know if this is correct.

The far edge of the box closest to the rifle is 142 pix @ 8.8 pix/in = 16.1 inches. The side edge of the box is being foreshortened by the cosine of the angle that the top of the box makes with the film plane. But it is also closer to the camera so we can't use 8.8 pix/in to measure it. However, the ratio of the image width of the far edge of the box / the width of the near edge gives us the pixels/inch for the near side: 156px / 142px = 1.09 * 8.8px/in = 9.7px/in. The foreshortening correction is then the average pixels/inch from far to near = (8.8 + 9.7) / 2 = 9.25 pixels/inch. The foreshortening correction for the box top is 170px * 1.09 = 185.3px. The length of the box is then 185.3px / 9.25px/in =20.0 inches

If my original estimate of 34 inches for the the MC without the barrel is accurate, then the dimensions of the box is 16" x 20" +/- .5".
« Last Edit: February 13, 2021, 02:10:48 AM by Jack Trojan »

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Re: Roger Craig
« Reply #352 on: February 13, 2021, 02:09:27 AM »


Offline Walt Cakebread

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Re: Roger Craig
« Reply #353 on: February 13, 2021, 02:33:52 AM »
I won't draw this out for you, however, you need to measure the total length of 1 of your MCs from the end of the stock to the end of the barrel. The image of the MC is 332 pixels, which = X inches. When I measured the MC minus the barrel I got 300 pixels / 34 inches = 8.8 pixels/inch. This tells me that a rifle with the barrel should be 37.7 inches. Let me know if this is correct.

The far edge of the box closest to the rifle is 142 pix @ 8.8 pix/in = 16.1 inches. The side edge of the box is being foreshortened by the cosine of the angle that the top of the box makes with the film plane. But it is also closer to the camera so we can't use 8.8 pix/in to measure it. However, the ratio of the image width of the far edge of the box / the width of the near edge gives us the pixels/inch for the near side: 156px / 142px = 1.09 * 8.8px/in = 9.7px/in. The foreshortening correction is then the average pixels/inch from far to near = (8.8 + 9.7) / 2 = 9.25 pixels/inch. The foreshortening correction for the box top is 170px * 1.09 = 185.3px. The length of the box is then 185.3px / 9.25px/in =20.0 inches

If my original estimate of 34 inches for the the MC without the barrel is accurate, then the dimensions of the box is 16" x 20" +/- .5".

Thanks Jack....I wanted the dimensions of the box to use as a scale for the rifle....

This tells me that a rifle with the barrel should be 37.7 inches. Let me know if this is correct.

The official manufacturers length of a model 91/38 carcano short rifle is: 40.125 inches. ( from a gun manual)  And that's exactly what my carcano measures.

You have reversed my thinking....I suspect the photo of rifle is a  composite (ie; a fake photo which seems to show the carcano) but in reality the rifle in the  photo is a created composite.

Offline Walt Cakebread

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Re: Roger Craig
« Reply #354 on: February 13, 2021, 05:20:37 AM »
Thanks Jack....I wanted the dimensions of the box to use as a scale for the rifle....

This tells me that a rifle with the barrel should be 37.7 inches. Let me know if this is correct.

The official manufacturers length of a model 91/38 carcano short rifle is: 40.125 inches. ( from a gun manual)  And that's exactly what my carcano measures.

You have reversed my thinking....I suspect the photo of rifle is a  composite (ie; a fake photo which seems to show the carcano) but in reality the rifle in the  photo is a created composite.

John Iocoletti  posted the evidence list that lists the dimensions of the boxes, but i doubt that John would post that information, if he thought that he was aiding me.

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Re: Roger Craig
« Reply #354 on: February 13, 2021, 05:20:37 AM »


Offline Walt Cakebread

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Re: Roger Craig
« Reply #355 on: February 13, 2021, 04:44:32 PM »
I won't draw this out for you, however, you need to measure the total length of 1 of your MCs from the end of the stock to the end of the barrel. The image of the MC is 332 pixels, which = X inches. When I measured the MC minus the barrel I got 300 pixels / 34 inches = 8.8 pixels/inch. This tells me that a rifle with the barrel should be 37.7 inches. Let me know if this is correct.

The far edge of the box closest to the rifle is 142 pix @ 8.8 pix/in = 16.1 inches. The side edge of the box is being foreshortened by the cosine of the angle that the top of the box makes with the film plane. But it is also closer to the camera so we can't use 8.8 pix/in to measure it. However, the ratio of the image width of the far edge of the box / the width of the near edge gives us the pixels/inch for the near side: 156px / 142px = 1.09 * 8.8px/in = 9.7px/in. The foreshortening correction is then the average pixels/inch from far to near = (8.8 + 9.7) / 2 = 9.25 pixels/inch. The foreshortening correction for the box top is 170px * 1.09 = 185.3px. The length of the box is then 185.3px / 9.25px/in =20.0 inches

If my original estimate of 34 inches for the the MC without the barrel is accurate, then the dimensions of the box is 16" x 20" +/- .5".

Thank you...I'm happy that you're taking a different approach to the solution....

I used a known measurement from my Carcano and compared it to that measurement in the photo, and found that the multiplier for the photo is 16.29.    IOW....The box would be 18.7 inches wide.... and the rifle would be 43.5 inches long.

I really need the actual measurement of the box......
« Last Edit: February 13, 2021, 06:23:01 PM by Walt Cakebread »

Offline Jack Trojan

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Re: Roger Craig
« Reply #356 on: February 14, 2021, 01:50:20 AM »
Thank you...I'm happy that you're taking a different approach to the solution....

I used a known measurement from my Carcano and compared it to that measurement in the photo, and found that the multiplier for the photo is 16.29.    IOW....The box would be 18.7 inches wide.... and the rifle would be 43.5 inches long.

I really need the actual measurement of the box......

The box dimensions are standard 16" x 20" x 14". But you need a ruler of known dimensions within the image to confirm it. You've got it backwards trying to use the box to measure the MC. You must use the known dimensions of the MC to measure the box. Only if the box resolves to a standard size can you use it as a ruler. This will in turn confirm the dimensions of the MC.

Not sure if the image files I linked to will show since they don't for me. My approach is applying photogrammetry which is just measuring 3D objects in 2D photos. Our unit of measurement for images will be pixels which we must convert to a physical unit such as inches. I composed the following image to help you out:



I isolated the areas we intend to measure, doubled the resolution and brightened it up enough to clearly make out the ends of the MC. Fortunately, the rifle is almost orthogonal to the POV so there will be no foreshortening to consider. This occurs when one end of an object is farther away from the camera than the other end which compresses the image. The image length is reduced by the cosine of the angle of the lean away from (or towards) the camera. In this case the lean is negligible so we can take direct measurements of the rifle on the image.

In the image I posted I took measurements from the middle of the buttend to the end of the barrel, to the sights, to the piece that Oswald supposedly smuggled into the TSBD in a paper bag (34"), and to the end of the action at the forend. Then I measured the width of the far end of the box.

Note that I cut and pasted horizontal strips of the measurements  at the upper left corner of the image so you can count the pixels for each measurement. Copy and paste this image into MS Paint and hold your cursor at the end of each segment and note the pixel x,y (bottom left corner of Paint). Also note that I placed the measurement of the box as the bottom segment which I denoted with a red bullseye. That is the pixel width for the box. Because it is so close to the rifle, we can assume that both objects occupy the same plane and are commonly scaled.

For example: The rifle part without the barrel which Oswald supposedly smuggled into the TSBD is thought to be 34". In my prev image, that measurement on the MC is 894 pixels long.



This segment is 894 pixels / 34 inches = 26.3 pixels/inch (specific to this image). The box width is 426 pixels (red bullseye) @ 26.3 pixels/inch = 16.2 inches. This tells me that the 34" measurement of the barrelless MC is accurate.

Now it's your turn to measure these lengths (inches) on your own MC for the segments on the image I posted, and calculate the pixels/inch ratios. They should ALL be 26.3 pixels/inch to be an authentic MC as portrayed in the image. Let me know if the rifle matches yours. Otherwise, forget about the box.

« Last Edit: February 14, 2021, 01:58:41 AM by Jack Trojan »

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Re: Roger Craig
« Reply #356 on: February 14, 2021, 01:50:20 AM »


Offline Walt Cakebread

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Re: Roger Craig
« Reply #357 on: February 14, 2021, 02:52:34 AM »
The box dimensions are standard 16" x 20" x 14". But you need a ruler of known dimensions within the image to confirm it. You've got it backwards trying to use the box to measure the MC. You must use the known dimensions of the MC to measure the box. Only if the box resolves to a standard size can you use it as a ruler. This will in turn confirm the dimensions of the MC.

Not sure if the image files I linked to will show since they don't for me. My approach is applying photogrammetry which is just measuring 3D objects in 2D photos. Our unit of measurement for images will be pixels which we must convert to a physical unit such as inches. I composed the following image to help you out:



I isolated the areas we intend to measure, doubled the resolution and brightened it up enough to clearly make out the ends of the MC. Fortunately, the rifle is almost orthogonal to the POV so there will be no foreshortening to consider. This occurs when one end of an object is farther away from the camera than the other end which compresses the image. The image length is reduced by the cosine of the angle of the lean away from (or towards) the camera. In this case the lean is negligible so we can take direct measurements of the rifle on the image.

In the image I posted I took measurements from the middle of the buttend to the end of the barrel, to the sights, to the piece that Oswald supposedly smuggled into the TSBD in a paper bag (34"), and to the end of the action at the forend. Then I measured the width of the far end of the box.

Note that I cut and pasted horizontal strips of the measurements  at the upper left corner of the image so you can count the pixels for each measurement. Copy and paste this image into MS Paint and hold your cursor at the end of each segment and note the pixel x,y (bottom left corner of Paint). Also note that I placed the measurement of the box as the bottom segment which I denoted with a red bullseye. That is the pixel width for the box. Because it is so close to the rifle, we can assume that both objects occupy the same plane and are commonly scaled.

For example: The rifle part without the barrel which Oswald supposedly smuggled into the TSBD is thought to be 34". In my prev image, that measurement on the MC is 894 pixels long.



This segment is 894 pixels / 34 inches = 26.3 pixels/inch (specific to this image). The box width is 426 pixels (red bullseye) @ 26.3 pixels/inch = 16.2 inches. This tells me that the 34" measurement of the barrelless MC is accurate.

Now it's your turn to measure these lengths (inches) on your own MC for the segments on the image I posted, and calculate the pixels/inch ratios. They should ALL be 26.3 pixels/inch to be an authentic MC as portrayed in the image. Let me know if the rifle matches yours. Otherwise, forget about the box.

Since you seem to be very good at analyzing photos.....Here's the deal....  I don't believe that the rifle in the photo was actually there.

1st off ---I believe the photo was taken at night..... The rifle was discovered at 1:22pm and the in situ photo was allegedly taken before 1:45pm , so the area should be bathed in bright sunshine from the window that was in the west wall.

2nd --- If the photo was taken at night then the carcano wouldn't have been available , because the FBI took custody of it. and it was in Wash DC

3rd--- The stock on the rifle does not look like the stock of a carcano but the muzzle does look like a carcano.

I wonder if they could have used a mauser ( the stock looks like a mauser)  to take the photo and then altered the photo by placing the photo of the muzzle of a carcano ( we know they had many photos of the carcano.) on the photo.

Offline Jack Trojan

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Re: Roger Craig
« Reply #358 on: February 14, 2021, 10:30:21 PM »
Since you seem to be very good at analyzing photos.....Here's the deal....  I don't believe that the rifle in the photo was actually there.

1st off ---I believe the photo was taken at night..... The rifle was discovered at 1:22pm and the in situ photo was allegedly taken before 1:45pm , so the area should be bathed in bright sunshine from the window that was in the west wall.

2nd --- If the photo was taken at night then the carcano wouldn't have been available , because the FBI took custody of it. and it was in Wash DC

3rd--- The stock on the rifle does not look like the stock of a carcano but the muzzle does look like a carcano.

I wonder if they could have used a mauser ( the stock looks like a mauser)  to take the photo and then altered the photo by placing the photo of the muzzle of a carcano ( we know they had many photos of the carcano.) on the photo.

1) The shadows in the photo are very pronounced and were not created from the camera flash. The light source came from the right of frame either from the sun or lighting on the 6th floor. If you had a surveyed overhead image of the TSBD (Google Maps) and knew the layout of the 6th floor relative to North, and knew the sun angle at 1:45pm on that fateful day (which any astronomy program can give you) then you could tell whether the shadows contradict the timeline. But that's way too much work, and who would believe you? But fill your boots.

2) Yes, unless the rifle was photographed when the DPD claims they are busted..again, just add it to the plethora of other shenanigans they were up to. How come Alyea wasn't allowed to film the rifle in situ? Because Fritz called the shots.

3) I agree that the stock does not match the MC but the forend definitely matches the MC. If you rotate the rifle Day is holding up ~90 degrees along its axis so it is perpendicular to the floor, it's a perfect match, which includes the 2 screws.



However, I couldn't make the rifles match over their entire length when I scaled the forends to match. I couldn't attribute the discrepancies to a simple rotation or foreshortening.

You should be able to settle this by setting your MC down in a similar orientation and take some pics from the same POV and compare the images. You should be able to either match the in situ photo exactly or the gun is smoking. Then you could answer all your own questions. That is unless you don't actually have a MC. Then I would understand your reluctance.

Considering how Fritz controlled the crime scene and exactly how it got documented, including staging an in situ photo of the 3 hulls, which he removed from his pocket and tossed onto the floor in the sniper's nest and had a rookie cop take the photo, I wouldn't doubt that the photo of the MC ditched neatly beside the box is just more Fritz bullspombleprofglidnoctobunse. Add it to the list. Otherwise, I'm done with this one until you post your re-enactment photo with your MC.

Good luck!




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Re: Roger Craig
« Reply #358 on: February 14, 2021, 10:30:21 PM »


Offline Walt Cakebread

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Re: Roger Craig
« Reply #359 on: February 14, 2021, 11:51:01 PM »
1) The shadows in the photo are very pronounced and were not created from the camera flash. The light source came from the right of frame either from the sun or lighting on the 6th floor. If you had a surveyed overhead image of the TSBD (Google Maps) and knew the layout of the 6th floor relative to North, and knew the sun angle at 1:45pm on that fateful day (which any astronomy program can give you) then you could tell whether the shadows contradict the timeline. But that's way too much work, and who would believe you? But fill your boots.

2) Yes, unless the rifle was photographed when the DPD claims they are busted..again, just add it to the plethora of other shenanigans they were up to. How come Alyea wasn't allowed to film the rifle in situ? Because Fritz called the shots.

3) I agree that the stock does not match the MC but the forend definitely matches the MC. If you rotate the rifle Day is holding up ~90 degrees along its axis so it is perpendicular to the floor, it's a perfect match, which includes the 2 screws.



However, I couldn't make the rifles match over their entire length when I scaled the forends to match. I couldn't attribute the discrepancies to a simple rotation or foreshortening.

You should be able to settle this by setting your MC down in a similar orientation and take some pics from the same POV and compare the images. You should be able to either match the in situ photo exactly or the gun is smoking. Then you could answer all your own questions. That is unless you don't actually have a MC. Then I would understand your reluctance.

Considering how Fritz controlled the crime scene and exactly how it got documented, including staging an in situ photo of the 3 hulls, which he removed from his pocket and tossed onto the floor in the sniper's nest and had a rookie cop take the photo, I wouldn't doubt that the photo of the MC ditched neatly beside the box is just more Fritz bullspombleprofglidnoctobunse. Add it to the list. Otherwise, I'm done with this one until you post your re-enactment photo with your MC.

Good luck!

The shadows in the photo are very pronounced and were not created from the camera flash. The light source came from the right of frame either from the sun or lighting on the 6th floor.

The shadows in the photo are very pronounced and were created by a bright light, ( the camera flash)  The light source came from the right of frame 

The photo was taken from the south of the rifle...The rifle is pointed directly east .....so north is away from the camera toward the top of the photo. ( this is verified by the map that Studebaker drew of the sixth floor , He designated the photo DP#22 and placed the arrow on that sight with the camera  pointing due north) The shadows appear to have been created by the camera flash....AND they are being cast slightly east from directly overhead, and since the sun wasn't inside the TSBD the shadow was not created by the sun. And the lighting inside the building was barely adequate.... ( There are many photos that show the old fashioned lighting)  However this location is only about three feet from the West window that would have had bright sunshine blazing through it at 1:45 pm.

I couldn't make the rifles match over their entire length when I scaled the forends to match. I couldn't attribute the discrepancies to a simple rotation or foreshortening.

I also cannot make the rifle in the photo match the actual true length ( 40.125")of a model 91/38 carcano... And I've used the muzzle end of the rifle in the photo as a scale.   The rifle in the photo is over two inches longer than a model 91/38....