Ray Intersecting Rectangle

[Home]   [Puzzles & Projects]    [Delphi Techniques]   [Math topics]   [Library]   [Utilities]



Search WWW

Search DelphiForFun.org

As of October, 2016, Embarcadero is offering a free release of Delphi (Delphi 10.1 Berlin Starter Edition ).     There are a few restrictions, but it is a welcome step toward making more programmers aware of the joys of Delphi.  They do say "Offer may be withdrawn at any time", so don't delay if you want to check it out.  Please use the feedback link to let me know if the link stops working.


Support DFF - Shop

 If you shop at Amazon anyway,  consider using this link. 


We receive a few cents from each purchase.  Thanks


Support DFF - Donate

 If you benefit from the website,  in terms of knowledge, entertainment value, or something otherwise useful, consider making a donation via PayPal  to help defray the costs.  (No PayPal account necessary to donate via credit card.)  Transaction is secure.

Mensa Daily Puzzlers

For over 15 years Mensa Page-A-Day calendars have provided several puzzles a year for my programming pleasure.  Coding "solvers" is most fun, but many programs also allow user solving, convenient for "fill in the blanks" type.  Below are Amazon  links to the two most recent years.

Mensa 365 Puzzlers  Calendar 2017

Mensa 365 Puzzlers Calendar 2018

(Hint: If you can wait, current year calendars are usually on sale in January.)


Feedback:  Send an e-mail with your comments about this program (or anything else).

Search DelphiForFun.org only



Problem Description

Here's the email which prompted this program:

Contact requested
I have a puzzle that's bugged me for a long time. This is probably just my failure to learn enough math.

|         |       |             
E ---- F---G
|         |       |

Given a bounding rectangle defined by points A,B,C,D;. Point F is a point in the absolute middle of the bounding rectangle, given half way between AB and BC. Point E is half way between A and D. Point G is half way between B and C. The last variable is an angle THETA from 0-360 that is based upon the line FG, where 0/360 is defined by the line of FG and 180 is defined by FE.

The puzzle is how to create a formula to define a line that will always intercept point F using angle THETA, but will also define the two X,Y coordinate points where the line will intercept the perimeter of the bounding
rectangle. For instance, if angle THETA was 0, 180 or 360, the line would intercept at points F and G.

If this is an easy one for people.. I'm gonna go back to school. All my solutions had some random error, but it might be due to my need to convert back & forth from radians.

Background & Techniques

 Here's the derivation of an algorithm that will provide the answer.  First a more concise statement of the problem:

Determine which side or sides of a given rectangle are intersected by a ray from the center at a given angle.

Assume that the center of rectangle ABCD is at coordinates (0,0) and let
S1=length(AB)/2 and
S2=length(BC)/2 and
the ray extends from (0,0) at angle theta (q), intersecting the rectangle at distance R

Then the equations for the lines containing the sides can be defined as follows:
For AB: Y=S2,  with -S1<=X<=S1
for BC: X=S1 with -S2<=Y<=S2,
for CD: Y=-S2, -S1<=Y<=S1 and
for DA: X=-S1, -S2<=Y<=S2

The intersection coordinates for the ray line at angle q can be defined as X=R*cos(q) and Y=R*sin(q) with  R being  the distance from the center at (0,0) to the point of intersection.

Let's derive the intersection point for the ray and side AB
Y=S2 and because Y=R*sin(q), we can divide and define R as R=S2/sin(q).
Substituting for R in the ray equation for X we have X=(S2/sin(q))*cos(q)) or by rearranging terms
X=S2*(cos(q)/sin(q))Cos(q)/sin(q) is the definition of cotangent(theta) so X can also be caculated as X=S2*cotan(q)

The above equations do not take into account that we are checking a line segments, not lines of unlimited length, so we need the  extra checks to make sure that the intersection point is within the bounds of AB (Y is between
-S1 and +S1) and that we are not talking about the "backend" of the ray (angle must be between -90 (or 270) and +90 degrees. And we need to detect and not try to evaluate the special case when the ray happens to be
parallel to AB. It will never intersect and the cotangent function is undefined (because the divisor, sin(q), has
value 0.

Similar analyses can be applied to the other three sides to determine the point or points of intersection for the ray.  It will intersect 2 adjacent sides if it happens to pass exactly through a corner of the rectangle.

The sample program allows you to set the lengths of the sides of the rectangle and the angle of the ray.  It draws an approximate image of the resulting figure and displays which side or sides are intersected by the ray.

Notes for Programmers

The code to derive the intersection coordinates is a straightforward implementation of the above description.  The main addition is to draw the diagram to visually show the rectangles and the ray as the parameters are changed by the user.   A procedure DrawRect is called which uses a TImage canvas to redraw the rectangle, label the corners and draw a line from the center to a point just out side of the rectangle.  A common OnChange exit, ValueChange, is called each time one of the 3 TSpinedit values (ABlength, CDlength, ThetaDegrees) changes.

There is a bug, or at least  deficiency, in Delphi's TSpinedit code when typing rather than using the up-down arrows is used to change the value.  It attempts to convert the blank text field to an integer using Try-Except logic.  This works OK unless we are running in Design mode when the EConvertErrors interrupt the execution to report the error.   It "bugs" me, so I check for a blank Text property and exit if true at the  start of the ValueChange procedure.  

One more programmer's trick at the beginning of ValueChange.   The initial text describing the program is initially displayed in the same TMemo where the intersection results are displayed after each parameter change.  On initial entry I want to display the results without clearing that introductory text.  I do that by calling ValueChange in the FormActivate method, but checking the Sender parameter at the beginning to see who caller was and only clearing the TMemo when the caller was one of the TSpinedit controls.       

Running/Exploring the Program 

bullet Download source
bullet Download  executable

Suggestions for Further Explorations

Not much that I can think of.  Perhaps  allow the user to click and drag the ray to rotate it with the mouse.  Nah, too much coding effort for the benefit J.
Displaying the X-Y axes and (X,Y) coordinates for the origin, corners, and intersection point would enhance the visual display .


Original Date: October 16, 2007 

Modified: July 29, 2017

  [Feedback]   [Newsletters (subscribe/view)] [About me]
Copyright 2000-2017, Gary Darby    All rights reserved.