Re: [CR]Appeal for Legnano Decal Help

(Example: Books:Ron Kitching)

Date: Mon, 15 Jan 2007 14:39:19 -0800 (PST)
From: Donald Gillies <gillies@cs.ubc.ca>
To: classicrendezvous@bikelist.org
Subject: Re: [CR]Appeal for Legnano Decal Help

With a picture from a bicycle tube you usually have known dimensions. Using these dimension you can scan and import the picture into a CAD program like autocad, MS-Visio, or Adobe Illustrator, and then scale up the picture to match the grid on the background of the cad surface.

Once the picture is properly imported and scaled, you can take pecise measurements of the decal using the software (subject to the fact that it is wrapped on a tube and then projected onto a flat screen.)

For example, you can draw lines and use the "show properties" menu of the line to measure things in the picture. So if the decal wraps around the tube and has been properly scaled, you should be able to sketch a line and measure the line to find the height of the decal. As for the width of the decal - assumning that it wraps width-wise around the downtube - you'll need to measure some width of the decal (100%, 50%, 25%) in inches and then use trig to calculate the overall width of the decal.

Consider this, assume i'm looking at a traditional CAMPAGNOLO decal that's placed right on the TOP TUBE (to make calculations easy) :

+vv^v^v^V^V^V^v+ | | | +-------+ | ^ | | DECAL | | | height | +-------+ | v | | +v^v^vv^v^v^v^^+

<===== 1" =====> flattened

<=== 1.578" ===> actual = 180 degrees

you know the flat_width of the tube is 1" (in any picture) the overall circumference of the tube is therefore 3.14159 ... inches the overall actual_width is therefore 1.578 ... inches total

tube radius (r) must be 0.5" on a 1" tube.

+-- DECAL_WIDTH --+

__----__ _- -_ _- - / \ /| \ / |\ / | \ alpha / | \ / | \ / | \ | | theta1\ /theta2 | | =============+============= left right decal decal extent extent <= 0.33 =><= 0.33 =>

decal_width = left_decal_extent + right_decal_extent

let's say :

left_decal_extent = 0.33 in the picture right_decal_extent = 0.33 in the picture

r cos theta = x 0.5 * cos(theta1) = 0.33 ; theta1 = inv_cos(0.66) = 48 degrees 0.5 * cos(theta2) = 0.33 ; theta2 = inv_cos(0.66) = 48 degrees

alpha + theta1 + theta2 = 180 alpha = 180 - 48 - 48 = 180 - 96 = 84 degrees decal = (84 / 180) * 1.578 = 0.74 inches wide, actually.

- Don Gillies
San Diego, CA, USA