Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

It's not as easy as it sounds. Basically I can move $A$ and check the distance to $B$. However there is a $\pm0.015$ error range , in other words if the real distance is $0.29$, it will get rounded to $0.3$. So the possible values are : $0.03$, $0.06$, $0.09$ and so on

I tried drawing two circles and finding the intersection but the thing in the square root is negative. I assume this is because of the rounding - the circles don't actually overlap.

Is there a method to find the coords of $B$ given the error margin? (note the coords need to be accurate in a circle of $0.05$ around the real coords) I can move $A$ at any point I want and get the rounded distance - can I increment until the value changes and derive the actual real distance from that?

share|improve this question

1 Answer 1

For each point $A$ that you choose, you get a ring of thickness 0.3. Your problem is that even if two such rings intersect, their approximating circles don't always intersect.

But this only happens if the two rings graze each other (either internally or externally). If each ring passes through the other ring completely, into its interior, then the approximating circles will always intersect (in two points).

To make sure this happens, you want to choose your first point $A_1$ to be far enough away from $B$ that the ring has a large interior -- say, a distance of 5 or more. Of course you don't know where $B$ is, but if your first choice is too close, just choose another point at a distance of 10 from the first-choice point.

Suppose the resulting distance is $d$. Now choose your second point $A_2$ at a distance $2d$ from $A_1$. If the distance from $B$ to this new point is too close to $d$ or $3d$, then the two rings will graze (or almost graze, which is bad because the intersecting regions are too long); so choose another point at the same distance from $A_1$, but in a direction at right angles to the original direction.

Now your two approximating circles will intersect cleanly in two points. To find out which of them approximates $B$, just choose your third point $A_3$ to be one of these two points.

share|improve this answer
    
Thanks Tony , but that would be way to complicated to implement. If it rounds up , can I find a real value of the distance? I.E if it skips from 0.03 to 0.06 then I just crossed 0.015 right? Is that true and can I then use the cosine rule or something to find the coords? I tried but it seems the resulting values do not form a triangle... –  Franken K Nov 13 '12 at 22:47
    
Way too complicated?! I give up... –  TonyK Nov 13 '12 at 22:49
    
Sorry if I haven't made it clear , I will need to program this. I'm a student and I've never done anything regarding rings and I have a bit of trouble understanding your method...Either way , thanks for your time. –  Franken K Nov 13 '12 at 22:59
    
I've given you an explicit algorithm. I urge you to take the trouble to understand it! Draw some pictures if it helps. –  TonyK Nov 13 '12 at 23:33

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.