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Calibration of the Restoration Photograph

For the following, I am referring to the full digital Shroud image publicly distributed by the Arcidiocesi di Torino, after the summer 2002 restoration. This image was taken from http://www.sindone.org/it/scient/restauro_gallery.htm and is copyrighted Arcidiocesi di Torino (The URL of this image was http://www.sindone.org/restauro/hires/sindone_recto.jpg). The image has 613px (pixel) in width and 2286px in length. By convention, the upper left corner has coordinates (0,0); consequently, the lower right corner has coordinates (612,2285).

During the 2002 restoration, various length measurements of the Shroud were taken by Karlheinz Dietz and Gian Maria Zaccone: 441.5 cm for the right length, and 442.5 cm for the left length. The bottom width is 113.0 cm and the top width is 113.7 cm. These values were reported in Sindone 2002, Mechthild Flury-Lemberg, 2003, editor ODPF -- they can be used to calibrate that image: one pixel is approximately 1.985 mm. The details of this calibration is explained in the following paragraph. Essentially, the width of the Shroud is 568 px; so 1 px = 1130/568 mm which is approximately 1.985 mm.

Note: when the photo was taken, the Shroud was not stiched to its backing cloth---it was therefore in a more "relaxed" state, so it is longer and wider than previous photographs (e.g. Enrie, 1931).

The Shroud left upper corner is at (15,21), the left lower corner is at (19,2266), the right lower corner is at (594,2273) and its right upper corner is at (594,18). Even these four measurements are approximated to +- 1px due to the fuzziness of the Shroud borders on the image. Needless to say, the Shroud is not perfectly rectangular as for example the upper edge is concave; near its center it goes to (x,27) for x around 270. To measure the length (a.k.a. height) and width of the Shroud, endpoint locations must be assumed. For the length, points (270,27) and (270,2263) are used, that is its length is 2237 px. For the width, points (32,1116) and (599,1116) are used, that is its width is 568px. Assuming that the Shroud width is 1130 mm, one horizontal pixel is 1130/568 mm, that is 1.989 mm +- 0.01mm according to the +-1px approximation. Assuming that the Shroud length is 4425 mm, one vertical pixel is 4425/2237 mm, that is 1.978 mm. Assuming that the image is indeed symmetrical, we may use the calibration of 1 px = 1.985 mm with an error of less than 0.01mm.

A length measurement can be done by using your favorite graphic software. (For example Gimp, free and available for different flavors of Unix and Windows.) Usually you would download the image and by using the "compass tool" (a.k.a. rule tool) measure the number of pixels between two points. For a vertical and horizontal measurement, the length is the number of pixels multiplied by the calibration. Or you can use the previous links to do online measurements.

For instance, a measurement of the width of the "entire face" (face encompassing the hair on both sides) could be from (260, 1299) to (367, 1299); that is 108 horizontal pixels. Therefore, the width is 108*1.985 mm = 21.4 cm (rounded). This measurement is still disputable, but at least stating the endpoints makes it reproducible.

Note: measurements of body parts on the image does not necessarily translate as measurements on a purported real body draped by the Shroud: the image could have been created while the cloth draped the body, creating some wraparound distortions, or the Shroud could have stretched since the image was formed, the body does not lay flat, and so on.

Some other examples of measurements: The height of the frontal body image from (311, 1224) to (311, 2256) is 1032 pixels, that is 1032*1.985 mm = 2.05 m. (We certainly cannot conclude that this is the height of the Man of the Shroud as this measurement does not take into account many other factors.) The empty space from (320,1132) to (320,1224) gives 18.3 cm. The back body image from (320,81) to (320,1132) gives 2.09 m. Actually, all these endpoints are disputable: the contour of the image is not very clear, in particular near the feet.

Acknowledgments

The Secundo Pia's digital photo was sent to me by Prof. Giulio Fanti. The Enrie's photo was taken from Ray Schneider's web page which was rescaled by him to 613 pixels in width.

Some Simple Calculations that Migh Have Big Consequences

If you have read this far you might enjoy the following (real) conversation around some simple calculations. The following is a transcript of an interchange between defence attorney Robert Blasier and FBI Special Agent Roger Martz on July 26, 1995, in the courtroom of the O.J. Simpson trial:

LOS ANGELES, CALIFORNIA; WEDNESDAY, JULY 26, 1995 1:56 P.M.

Q:         Can you calculate the area of a circle 
           with a five-millimeter diameter?
A:         I mean I could. I don't...math I don't ...
           I don't know right now what it is.
Q:         Well, what is the formula for the area of a circle?
A:         Pi R Squared
Q:         What is pi?
A:         Boy, you are really testing me. 2.12... 2.17...
Judge Ito: How about 3.1214?
Q:         Isn't pi kind of essential to being a scientist
           knowing what it is?
A:         I haven't used pi since I guess I was in high school.
Q:         Let's try 3.12.
A:         Is that what it is? There is an easier way to do...
Q:         Let's try 3.14. And what is the radius?
A:         It would be half the diameter: 2.5
Q:         2.5 squared, right?
A:         Right.
Q:         Your honor, may we borrow a calculator?
[pause]
Q:         Can you use a calculator?
A:         Yes, I think. 
Q:         Tell me what pi times 2.5 squared is.
A:         19
Q:         Do you want to write down 19? Square millimeters, right?
           The area. What is one tenth of that?
A:         1.9
Q:         You miscalculated by a factor of two, the size, the 
           minimum size of a swatch you needed to detect EDTA
           didn't you?
A:         I don't know that I did or not. I calculated a little
           differently. I didn't use this. 
Q:         Well, does the area change by the different method of
           calculation?
A:         Well, this is all estimations based on my eyeball. I
           didn't use any scientific math to determine it. 

—David Blatner, the joy of pi