Working out scan angle - help

[andy_con]

http://www.pangolin.com/userhelp/scanangles.htm

so going by pango's info

width is 53cm
distance is 2m

so 2x2 = 4

0.53 / 2 = 0.265

0.265 * 4 = 1.06

closest match is 1.0724 meaning the half scan angle is 47 degrees

double that and 94 degrees........

my scanners arent running at 94 degrees

what am i doing wong??

[loopee]

Refer to this thread for a calculator that trwalters001 created

http://www.photonlexicon.com/forums/...ead.php?t=7086

Troy

[andy_con]

thanks .

[buffo]

Andy;

You need to solve the left side of the top equation first (.5 * W/D) and then take the inverse Tan (Tan to the -1 power) of that. Then multiply by two and you've got your angle. In this case, I get 15 degrees.

However, there is a much easier way to calculate this. Use the scan angle table and set up a ratio... The scan angle tables are set up with a fixed distance of 100 units. So take your width divided by your distance and multiply by 100. Look up that number on the table (the blue numbers), and find the scan angle that is closest. You numbers were 53/200 * 100, which is 26.5. That's close to 26.3, which is the number for 15 degrees!

Get it?

Adam

PS: I agree that Tim's calculator is also very easy to use, but the ratio method works anytime - even if you happen to be away from your home computer and don't have his tool installed, you can still get a number quickly.

[andy_con]

cheers guys .

[Xytrell]

You could also calculate it like you calculate divergence. 53cm at 2m is 265mm at 1m, and thus 0.265 radians (265mRad more familiarly). radians × 180 ÷ π = degrees. 0.265 ×180 ÷ π = 15.18°

Formula you can take with you (no computer or trigonometry required ):

A = 1.8S/πD

Where:
A is scan angle in degrees
S is size of projection in centimeters
D is distance of projection from projector in meters

[Lasermad]

Andy
you not got robs handy chart !

paul

[andy_con]

i forgot about that paul, till he mentioned it on the phone LOL