Finding coordinates of a point in a circle with angle

AI Thread Summary
To find the coordinates of a point on a circle given an angle and distance, the formula (x_?, y_?) = (x + d*cos(a), y + d*sin(a)) is used. The discussion highlights a coding issue where the calculated points were incorrect, leading to confusion about the angle measurements. It was discovered that the problem stemmed from using radians instead of degrees in the calculations. The user expressed gratitude for the clarification, noting the importance of checking measurement units in programming. This exchange emphasizes the common pitfalls in trigonometric calculations and the significance of unit consistency.
AlicanC
Messages
3
Reaction score
0
I have a point (x,y), a distance (d) and an angle (a). I need a method to find a point using x,y,d and a. For example:
http://c.imagehost.org/0836/03082008299.png
I will apply this on a 3D environment, but I am not going to use the 3rd dimension so it's safe.
This is also about programming, but I am capable of applying methods in the programming language so I asked it here.
 
Last edited by a moderator:
Mathematics news on Phys.org
<br /> (x_?,y_?) = (x+d\cos\alpha,y+d\sin\alpha)<br />
 
Thanks for the answer, but I have a problem. I wrote some code to put this in action. The code starts from "0" and adds "20" to the angle until the vaule is greater than to "360". I don't know why but the code produced this:
http://c.imagehost.org/0346/2008-08-03_092702.png
As you can see there are some mistakes. All are getting repeated but the ones I marked. Here is the debug output I get:
(1) Center of the circle located at (-49.013168, -1082.446166)
(1) Drawing at (50.986831, -1082.446166) Angle is 0.000000
(1) Drawing at (-8.204959, -991.151611) Angle is 20.000000
(1) Drawing at (-115.706977, -1007.934875) Angle is 40.000000
(1) Drawing at (-144.254455, -1112.927246) Angle is 60.000000
(1) Drawing at (-60.051891, -1181.835083) Angle is 80.000000
(1) Drawing at (37.218719, -1133.082763) Angle is 100.000000
(1) Drawing at (32.404930, -1024.385009) Angle is 120.000000
(1) Drawing at (-68.794525, -984.422180) Angle is 140.000000
(1) Drawing at (-146.576110, -1060.503662) Angle is 160.000000
(1) Drawing at (-108.859176, -1162.561401) Angle is 180.000000
(1) Drawing at (-0.294399, -1169.775878) Angle is 200.000000
(1) Drawing at (50.595352, -1073.606323) Angle is 220.000000
(1) Drawing at (-16.435035, -987.901672) Angle is 240.000000
(1) Drawing at (-122.032585, -1014.122192) Angle is 260.000000
(1) Drawing at (-141.187133, -1121.227050) Angle is 280.000000
(1) Drawing at (-51.222831, -1182.421752) Angle is 300.000000
(1) Drawing at (41.357345, -1125.261718) Angle is 320.000000
(1) Drawing at (26.953659, -1017.415100) Angle is 340.000000
(1) Drawing at (-77.382278, -986.554565) Angle is 360.000000
As you can see some values are too close. I am really bad at trigonometry, so I can't find the problem.
 
Last edited by a moderator:
The center is at (-49, -1082) and your first point, with the angle, 0 is at
(51, -1082) then radius, d, is 100.

So, when the angle is 20 the point should be
100(\cos(20)) + -49 \approx 45
100(\sin(20))+ -1082 \approx -1048

But you have:

Drawing at (-8.204959, -991.151611) Angle is 20.000000

*Could you be in radians instead of degrees?
 
futurebird said:
Could you be in radians instead of degrees?
Ah, you are right! I was making calculations with radians.

Thank you both so much! I was trying to figure this out for two days! Normally I post my questions to forums related to coding or just ask my friends, but can't get an answer :D
 
futurebird said:
Could you be in radians instead of degrees?
Always a problem if you forget. When I get errors like that, the first thing I do is to check the mode on my calculator. Lol.
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top