Roger Craig

[Walt Cakebread]

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?

Can you tell me the dimensions of the box above the rifle in the photo?  I estimate the width of the box at approximately 15 inches...But I know the FBI gave the dimensions of the boxes in the evidence list.   I just haven't found those dimensions yet.

[Jack Trojan]

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".

[Walt Cakebread]

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.

[Walt Cakebread]

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.

[Walt Cakebread]

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......

[Jack Trojan]

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.

[Walt Cakebread]

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.