If you can scan pixels, why not just count # of pixels in the arc and multiply by size of each pixel?
finding line intersection seems like a hard thing especially on a scanned image like this one. oh wait minute, it's an image... do you mean you want to find the ratio of arc to circumference? b.c. you can easily display a different sized image on screen, so there is no real absolute length. On Jan 3, 11:53 pm, "James Fang" <[EMAIL PROTECTED]> wrote: > It’s possible. > > Scan the digital picture pixel by pixel and line by line until you find the > first line with two black pixels ( the adjacent pixels is counted as only one > ),name the two pixels a1(xa1,ya1) and b1(xb1,yb2), continue scaning and you > will get a2(xa2,ya2) and b2(xb2,yb2), a3(xa3,ya3) and b3(xb3,yb3). > > If (ya2-ya1)/(xa2-xa1) = (ya3-ya2)/(xa3-xa2) , then a1 is the first > intersection. Store the coordenate of a1. > > Otherwise, b1 is the first intersection. > > Continue scanning, you will surely find the second intersection and the > coordinate of the centre of the circle. > > Since you got the 3 point of the triangle, you can caculate the theta angle > now. > > And finally, caculate the arc length by PI*r*r*(theta/360) > > _____ > > 发件人: [email protected] [mailto:[EMAIL PROTECTED] 代表 Daniel Bastidas > 发送时间: 2008年1月4日 10:19 > 收件人: [email protected] > 主题: [algogeeks] arc length > > Hi everybody. > > How can I find the arc length (L) in the picture attach if the only thing I > know is the radio of circumference. > I don´t know the coordenates of the intersection between line and > circumference neither the theta angle. > Any idea, it is possible...?. > Thanks --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected] To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/algogeeks -~----------~----~----~----~------~----~------~--~---
