Jump to content

  • Log In with Google      Sign In   
  • Create Account

We're offering banner ads on our site from just $5!

1. Details HERE. 2. GDNet+ Subscriptions HERE. 3. Ad upload HERE.

Simple beginner question (arctan, 2d angle stuff)

Old topic!
Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.

  • You cannot reply to this topic
3 replies to this topic

#1 Retro Owl   Members   -  Reputation: 169


Posted 11 January 2014 - 01:41 PM

I really suck at trigonometry, so this will be almost like an elementary school level question (hey, I'm a history major who's coding games for fun, what do you expect...).


My game uses 2d tile-based maps and at the moment I'm implementing a casting/shadowing/whatever's the name system so that tiles that aren't visible to the player (ones that are behind objects or walls) are not shown.


I have it otherwise done but I'm stuck with a very simple problem.


Basically what I need is a function that gives me the angle between two points (let's say point A and point B) on the map (I know there's no angle between two points, but you know what I mean). You tell the function the point coordinates x1, y1, x2 and y2 but it's very important that it should then return a value like this:


if Point B is directly above Point A, the function returns 0 (or 360)

if Point B is directly on the right side of Point A, the function returns 90

if Point B is directly below Point A, the function returns 180

If Point B is on Point A's left side, it returns 270.

(and in other cases it returns a value between these)


In short: I need a function that takes two points on the map (x1, y1, x2, y2) and returns a value that is converted into a number between 0 and 360, so that it follows the style presented above. I know I have to first count the delta values and I'm guessing I need to use Arctan or something, but I'm not sure how to do it and how to then convert the value into this 0-360 system that I'm looking for.


I wouldn't ask you to do my stuff for you, but I know this can be done in seconds when you know what you're doing. A simple function is enough, I'm sure I'll understand it once I see it, but feel free to give some extra pointers along the way.


If you need to use code (which you don't), please use pseudocode




PS. I promise to take a math class this year, if someone helps me.


#2 Brother Bob   Moderators   -  Reputation: 8454


Posted 11 January 2014 - 02:13 PM

Before I answer the question on how to convert your vector to an angle, I will ask; does your problem actually require angles, or are you just going to convert that angle to another vector later on or perhaps what you're after is really the sine or the cosine of the angle? More often than not, the angles themselves are not what you should be looking for, a better solution often exist involving linear algebra and no trigonometry.

#3 Retro Owl   Members   -  Reputation: 169


Posted 11 January 2014 - 03:34 PM

I would say that yes, my problems requires angles. I'm sure there's a better way to do all of this, but if someone can help me with the conversion I'm looking for, then at least I know I'll be able to get this feature to my game. Definitely not in the best possible way, but in a way that works for me.


I'm essentially a self-learned rogue programmer, so ultimately I only care about getting things to work, no matter how ugly the code is. I'm also at the moment 100% focused on actually coding my game, so learning new stuff that takes a while to swallow is not in the cards for me right now.


I will definitely later look at some line of sight and field of vision tutorials that were already recommended to me, but now I'd just like to get that one function that does the conversion for me. But naturally I welcome all extra tips that you can give me.

#4 Brother Bob   Moderators   -  Reputation: 8454


Posted 11 January 2014 - 04:13 PM

Some things are fundamental knowledge, and I would classify linear algebra as one. So while I agree that a working hackish solution is better than no solution if you just want to get your program working, linear algebra is something I would recommend learning as soon as possible.


But, that aside, the atan2 function is what you're after in this case. It gives you a result on the full circle instead of just half a circle for atan. Your description of the results is ambiguous, though. I will assume that "above" means "along the positive Y-axis" and "right" means "along the positive X-axis".


Given the vector D=B-A from point A to point B, the bearing of the vector D according to your description is atan2(D.x, D.y).

Old topic!
Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.