dpc wrote:If you know the length (L) of the green line segment between the red and yellow camera direction vectors in the diagram below, then the vertical distance between those two lines (i.e. the amount you'll need to translate the camera) is simply cos(cradle_angle / 2) * L. (The cradle_angle of the standard scanner is 90 deg)
dpc wrote:If you know the length (L) of the green line segment between the red and yellow camera direction vectors in the diagram below, then the vertical distance between those two lines (i.e. the amount you'll need to translate the camera) is simply L / cos(cradle_angle / 2). (The cradle_angle of the standard scanner is 90 deg)
Would you please be so kind to put your calculations into a spreadsheet?
In such a way that when you enter some of the measurements it will calculate the others/result.
Sure I will add up what i have been using when I get back to the house.
Can you explain which theorum of Pythagoras you have been using?
Have a look at Pythagorean Theorem to help you pick out the right one.
I used A^2 + B^2 = C^2 My triangle also had A and B sides with the same length and interior angle (isosceles)which would be 8.5^2 + 8.5^2 = 12^2 (the captureable area of the hypotenuse was only 11 inches however)
I have asked Rob to help out with the calculation of the sides of the cradle with only one side known. See:
Science! Or, a new page on the wiki for technical works by rob Â» 01 Apr 2011, 19:58
Thats awesome. It's hard to do since the other two sides could be any length except if the triangle is right or isosceles
I have some understanding of what you are explaining and have a theory about it at
Debate: camera positioning -- flexible vs. fixed
After I posted it I then found your excellent explanation. (Sorry I didn't finish reading the complete forum yet)
I will check it out. Thanks.
I am stuck because I never had any Trigonometry and it looks Rob and you did.
I actually haven't but I have been reviewing for the GRE. It has some geometry shortcuts. I stacked triangles to make a square. The square gave me a base of reference which was 17 inches vertical for an 11 inch book. I then added and subtracted book lengths / thicknesses based on that
Thank you in advance for your trouble.
dpc wrote:If you have a right triangle and know the length of one side and one of the other angles, you can find the length of the other sides using sine and cosine functions.
Given a right triangle with sides a, b, and c, with c being the hypotenuse, and BC being the angle formed by sides b and c:
a = sin(BC) * c
b = cos(BC) * c
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