MEDEVAC coverage

• Jul 21st 2009, 09:18 AM
x34903
MEDEVAC coverage
This is a fun problem that I am doing in my time off.

First, let me give you an intro. Helicopter medical evacuation (MEDEVAC) flies out of a hospital (H1), lands at the point of injury (P), and returns to H. This must be completed within X flying distance. Thus, the coverage of the MEDEVAC is said to be pi*(X/2)^2 around point H. Sometimes coverage areas overlap and we have to subtract the overlapping region. That is the easy part.

Now comes to the fun stuff. Let's say that a hospital (H2) is placed outside of the normal coverage area (somewhere between 0 and X from H1), but no MEDEVAC is placed there. This makes is a hospital that patients can be dropped off at, but helicopters coverage cannot be based out of. Thus, the helicopter must fly out of H1 and land at H2 within X distance. Now, we have to figure out the area of additional coverage due to the addition of H2 (but no additional helos).

I'm trying to do this in Matlab. I also would like to figure out a way to display all of this coverage some sort of a 2D graph.

The hardest part is explaining the problem.

I've been able to figure out the coverage area with just hospitals that have MEDEVAC capabilities. Once I throw the hospitals without the MEDEVAC capabilities, I don't even know where to begin to define the shape created by the additional coverage.

Since the MEDEVAC circles have equal radii, the following equations work for calculating the overlap between circles.
X = 100; %max distance MED aircraft can fly
r = X/2; %radius of the MED circle
d = 80; %distance from centers of two circles
q = 2*acos(d/(2*r)); %angle in radians
overlap = r^2*(q-sin(q)); %area of overlap
Area = 2*pi*r^2-overlap

I just need help moving on to calculating the area created by hospitals without MEDEVAC facilities.
• Jul 21st 2009, 06:59 PM
billa
I just started to read your problem, and I will continue to work on it, but I wanted to point something out.

I believe the non-Helicopter Hospitals add ellipse shapes to the 2d grid of coverage. The definition of an ellipse is that the sum of the distances from each point on it to two fixed foci is constant. One foci would be the helicopter-hospital, the other foci would be the non-heli-hospital. The distance would be 2X.

I don't have time to do anything else tonight, but this is a fun problem.
• Jul 22nd 2009, 03:33 AM
x34903
creating the ellipse is the trick
In the end it will be a problem of finding the intersection of an ellipse and a circle. I do agree that both hospitals will be foci (thus focus 1 will be the center of the circle). I just do not know how to create the equation for the ellipse given the requirement that the flight must be X length. Also, the eccentricity of the circle goes from 0 to 1 as the two foci move from 0 to X.

Also, once this problem is tackled, I run into the problem of using a larger coverage area that involves intersecting ellipses of variant eccentricities and rotation that share a common focus point.
• Jul 22nd 2009, 06:58 AM
billa
Quote:

I'm trying to do this in Matlab. I also would like to figure out a way to display all of this coverage some sort of a 2D graph.
Can you give me some information about how you are going to draw the shapes in MatLab ( does MatLab have any tools for drawing shapes ). I have heard of matLab but I don't know anything about it.

I don't think that finding a second degree equation for the ellipse given the coordinates of the hospitals would be hard, but I doubt matlab can draw an ellipse given an equation like this. In Java, for example, you draw ellipses by giving a bounding rectangle and rotating it.
• Jul 23rd 2009, 08:15 AM
x34903
It's been years since I graphed anything using Matlab, but it is a very powerful system. Once I get this problem solved, I plan on setting up the problem using cartesian coordinates and plotting the shapes. That part should be fairly easy. Any ideas on creating the ellipse using X as the max distance that the helicopter can fly? I can make it easier by using numbers...

X = 100; %max distance MED aircraft can fly
r = X/2; %radius of the MED circle
d = 80; %distance from centers of two circles
q = 2*acos(d/(2*r)); %angle in radians
overlap = r^2*(q-sin(q)); %area of overlap
Area = 2*pi*r^2-overlap

Thus, the helicopter can fly 100 miles and has a coverage area of 7,854 square miles. When helicopter-equipped hospitals are 80 miles apart, they have a total coverage area of 14,890 sqaure miles. What happens when the second hospital (H2) loses its helicopters (let's say due to maintenance)? Since all flights must come out of H1, the coverage around H2 decreases significantly.

Bottom line, the coverages becomes the area of the circle (7,854 sq mi) plus the area of the ellipse formed by the two foci (H1 and H2) minus the overlap.