Fix Keystoning via hardware adjustments

[the.traveller]

Jay,

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.

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 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 » 01 Apr 2011, 19:58

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 am stuck because I never had any Trigonometry and it looks Rob and you did.

Thank you in advance for your trouble.

[dpc]

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)

[jck57]

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)

A quick formula from pipefitting:

To find the length of the long side (hypotenuse) of a 45 degree triangle, multiply the short side length by 1.41.

[the.traveller]

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)

Thank you for the extra explanation. But I am not looking for the difference in height. (I still wil incorporate the above mentioned calculation into the spreadsheet)

I am looking to the 3 sides of your green triangle. I have to figure out the sides a and b if only c (the hypothenusa) is known. In my case I calculated that I need a board of 35 cm2 to hold my largest books. When I know what sides a and b are I can measure how large the total base should be.

So c = 35 cm2 ==> Hyphotenusa.
The angle of bc = 45 degrees,
ab = 90 degrees and
ac = 45 degrees ?????

When I look at this wikipage Calculating one side
But with these formulas you need to know at least 2 sides.

Because a2+b2= c2
a2+b2 = (35)2
a2+b2 = 1225

The formulas given are for 2 sides known.

Reading on I come upon Proof using similar triangles
When I look to the Proofs section it starts to boggle my mind because I never dealt with sinus and cosinus calculations.
AKA my brain locks up, to much to compute, to much to compute. to much to compute, to much.....

Is there a formula which can calculate the unknown sides a and b?

[the.traveller]

Thanks to Spamsickle who provided a calculation for a standard scanner which has a 90 degree cradle. I now can calculate how the cradle will look like. And from there the total size of the base.

See our discussion at Science! Or, a new page on the wiki for technical works

I made a spreadsheet with calculations which I will update each time I have a different calculation.
Because of the problem of finding the correct calculation method I will make a wiki page on the subject which will also have included the same spreadsheet. Anybody can put an explanation of his calculations in it for everybody to use if needed.

Not all of us have been studying Pythagorean theorem, or remember them clearly to use them easily. (like me )

Now I am trying to find a good formula to calculate the same but for a cradle which isn't 90 degrees. Say 120 degrees.
When I still know only one side, the hypothenusa = C, which will be the size of my largest book. Is there anybody who can tell how to calculate sides A and B. Where A being the short upstanding side and B the long horizontal plane.

We know that corner AB is still an 90 degrees angle, but BC will be (1/2*(180-120)) = 30 degrees and AC will be (120-(1/2*120)) = 60.
This is because AC and BC are opposite sides of an rectangle in which C is the hypothenusa.

[jay]

Jay,

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 » 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]

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

[the.traveller]

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

DPC I hope you can help with some further explanation.

How come if I make the calculation in Excel the answer for a is negative?

The respective corners are:
AB 90
BC 30
AC 60

For a being the short leg and b the long leg of the triangle.

If only 1 side is known and you need to find out

Side A -11,85637949
Formula:
a = sin(BC) * c

Side B 1,851017399
Formula:
b = cos(BC) * c

So showing each step
Fill in c 12
Fill in BC 30
sin(BC) -0,988031624
sin(BC) * c -11,85637949
cos(BC) 0,15425145
cos(BC) * c 1,851017399

However when I use the answers in the regular A2 + B2 = C2 formula, the answers are correspondent except the found value for a is now positive.

And there is another strange thing happening. In these calculations a suddenly becomes the long leg and b the short.

[rob]

You were using radians when you have to use degrees. Not sure how to set degrees in Excel, though.

[dpc]

Use the Excel function RADIANS to convert a value in degrees to radians.


sine of BC (30 deg) should be 0.5.

cosine of BC should be 0.8660254