How Contours Really Work
• Contours may LOOK like curves, but they are really just straight line segments.
How the contour is drawn depends on the value of the contour and the
values at the four corners!
grid[i][j] grid[i][j+1]
grid[i+1][j] grid[i+1][j+1]
if (CornersGreaterThanContour==1)… there are four possibilities:
nw ne
sw se
Possibility 1: Only the northwest corner is greater than the value of the contour.
if (CornersGreaterThanContour==1)… there are four possibilities:
nw ne
sw se
Possibility 2: Only the northeast corner is greater than the value of the contour.
if (CornersGreaterThanContour==1)… there are four possibilities:
nw ne
sw se
Possibility 3: Only the southeast corner is greater than the value of the contour.
if (CornersGreaterThanContour==1)… there are four possibilities:
nw ne
sw se
Possibility 4: Only the southwest corner is greater than the value of the contour.
if (CornersGreaterThanContour==2)… there are three possibilities:
nw ne
sw se
Possibility 1: The line should be drawn from the west edge to the east edge.
Either:
Both ne and nw are bigger than contour…
or
Both ne and nw are smaller than contour.
if (CornersGreaterThanContour==2)… there are three possibilities:
nw ne
sw se
Possibility 2: The line should be drawn from the north edge to the south edge.
Either:
Both ne and se are bigger than contour…
or
Both ne and se are smaller than contour.
if (CornersGreaterThanContour==2)… there are three possibilities:
nw ne
sw se
Possibility 3: Two contour lines pass through this box.
Either:
Both nw and se are bigger than contour…
or
Both nw and se are smaller than contour.
if (CornersGreaterThanContour==3)… there are four possibilities:
nw ne
sw se
Possibility 1: Only the northwest corner is less than the value of the contour.
if (CornersGreaterThanContour==3)… there are four possibilities:
nw ne
sw se
Possibility 2: Only the northeast corner is less than the value of the contour.
if (CornersGreaterThanContour==3)… there are four possibilities:
nw ne
sw se
Possibility 3: Only the southeast corner is less than the value of the contour.
if (CornersGreaterThanContour==3)… there are four possibilities:
nw ne
sw se
Possibility 4: Only the southwest corner is less than the value of the contour.
What will this program look like?
/* Compute CornersGreaterThanContour */
if (CornersGreaterThanContour == 0) { }
if (CornersGreaterThanContour == 1) { }
if (CornersGreaterThanContour == 2) { }
if (CornersGreaterThanContour == 3) { }
if (CornersGreaterThanContour == 4) { }
What will this program look like?
/* Compute CornersGreaterThanContour */
if (CornersGreaterThanContour == 0) {
/* Do nothing */
}
if (CornersGreaterThanContour == 1) { }
if (CornersGreaterThanContour == 2) { }
if (CornersGreaterThanContour == 3) { }
if (CornersGreaterThanContour == 4) {
/* Do nothing */
}
What will this program look like?
if (CornersGreaterThanContour == 1) { }
if (CornersGreaterThanContour == 2) { }
if (CornersGreaterThanContour == 3) { }
What will this program look like?
if (CornersGreaterThanContour == 1) {
if( nw > contour) { }
if( ne > contour) { }
if( se > contour) { }
if( sw > contour) { }
}
if (CornersGreaterThanContour == 2) { }
if (CornersGreaterThanContour == 3) { }
What will this program look like?
if (CornersGreaterThanContour == 2) { }
if (CornersGreaterThanContour == 3) { }
What will this program look like?
if (CornersGreaterThanContour == 2) {
if( (ne > contour && nw > contour) ||
(ne < contour && nw < contour)) { }
if( (ne > contour && se > contour) ||
(ne < contour && se < contour)) { }
if( (nw > contour && se > contour) ||
(nw < contour && se < contour)) { }
}
if (CornersGreaterThanContour == 3) { }
What will this program look like?
if (CornersGreaterThanContour == 3) {
if( nw < contour) { }
if( ne < contour) { }
if( se < contour) { }
if( sw < contour) { }
}
Deciding how to draw the contour
nw ne
sw se
Let’s say that you have determined that this is the kind of line you need to draw.
This line segment has a starting point and an ending point.
Deciding how to draw the contour
nw ne
sw se
What determines the location of the starting point?
INTERPOLATION!
Deciding how to draw the contour
nw ne
sw se
Suppose we are drawing the 1000 mb contour.
nw = 999.9
ne = 1022.1
Deciding how to draw the contour
nw ne
sw se
Suppose we are drawing the 1000 mb contour.
nw = 982.2
ne = 1001.1
Deciding how to draw the contour
nw ne
sw se
startlat = depends on:•latitude of nw•latitude of ne•value at nw•value at ne•value of contour
startlon = depends on:•longitude of nw•longitude of ne•value at nw•value at ne•value of contour
Deciding how to draw the contour
nw ne
sw se
interp(gridlatitude[i][j], &startlat, gridlatitude[i][j+1], nw, contour, ne);
interp(gridlongitude[i][j], &startlon, gridlongitude[i][j+1], nw, contour, ne);
Deciding how to draw the contour
nw ne
sw se
interp(gridlatitude[i][j+1], &endlat, gridlatitude[i+1][j+1], ne, contour, se);
interp(gridlongitude[i][j+1], &endlon, gridlongitude[i+1][j+1], ne, contour, se);
Deciding how to draw the contour
nw ne
sw se
Once you have (startlat, startlon) and (endlat, endlon):
1. tranform
2. clip
3. gline
The Steps
• Open the window
• Draw the base map
• Read the sao.cty file
• Read the current_sao.wxp file
• Produce grids of objectively analyzed data.
• Contour (like this….)
for(i=0;i<NUMROWS-1;i++) {for(j=0;j<NUMCOLS-1;j++) {
for(contour=mincon;contour<=maxcon;contour=contour+conint) {
CornersGreaterThanContour = ?????;
if (CornersGreaterThanContour==0) { }if (CornersGreaterThanContour==1) { }if (CornersGreaterThanContour==2) { }if (CornersGreaterThanContour==3) { }if (CornersGreaterThanContour==4) { }
}
}}
for(i=0;i<NUMROWS-1;i++) {for(j=0;j<NUMCOLS-1;j++) {
for(contour=mincon;contour<=maxcon;contour=contour+conint) {
CornersGreaterThanContour = ?????;
if (CornersGreaterThanContour==0) { }if (CornersGreaterThanContour==1) { }if (CornersGreaterThanContour==2) { }if (CornersGreaterThanContour==3) { }if (CornersGreaterThanContour==4) { }
}
}}
For every “square” on the map…
(Notice that there are
(NUMROWS-1)x(NUMCOLS-1) squares…)
for(i=0;i<NUMROWS-1;i++) {for(j=0;j<NUMCOLS-1;j++) {
for(contour=mincon;contour<=maxcon;contour=contour+conint) {
CornersGreaterThanContour = ?????;
if (CornersGreaterThanContour==0) { }if (CornersGreaterThanContour==1) { }if (CornersGreaterThanContour==2) { }if (CornersGreaterThanContour==3) { }if (CornersGreaterThanContour==4) { }
}
}}
For every possible contour level…
(We’ll discuss how to figure this out shortly.)
for(i=0;i<NUMROWS-1;i++) {for(j=0;j<NUMCOLS-1;j++) {
for(contour=mincon;contour<=maxcon;contour=contour+conint) {
CornersGreaterThanContour = ?????;
if (CornersGreaterThanContour==0) { }if (CornersGreaterThanContour==1) { }if (CornersGreaterThanContour==2) { }if (CornersGreaterThanContour==3) { }if (CornersGreaterThanContour==4) { }
}
}}
Determine the number of corners on this square
that are greater than the current contour level.
(This will take about 5 lines of code.)
for(i=0;i<NUMROWS-1;i++) {for(j=0;j<NUMCOLS-1;j++) {
for(contour=mincon;contour<=maxcon;contour=contour+conint) {
CornersGreaterThanContour = ?????;
if (CornersGreaterThanContour==0) { }if (CornersGreaterThanContour==1) { }if (CornersGreaterThanContour==2) { }if (CornersGreaterThanContour==3) { }if (CornersGreaterThanContour==4) { }
}
}}
For each of the 5 possible values of
CornersGreaterThanContour, you’ll need the
elaborate “if” statements discussed earlier.
for(i=0;i<NUMROWS-1;i++) {for(j=0;j<NUMCOLS-1;j++) {
for(contour=mincon;contour<=maxcon;contour=contour+conint) {
CornersGreaterThanContour = ?????;
if (CornersGreaterThanContour==0) { }if (CornersGreaterThanContour==1) { }if (CornersGreaterThanContour==2) { }if (CornersGreaterThanContour==3) { }if (CornersGreaterThanContour==4) { }
}
}}
Where do you get these values?
mincon, maxcon, conint
mincon, maxcon, conint
• You could just prompt the user for these three values.
• Better: prompt the user for conint, and compute mincon and maxcon!
• But, to do this, you first need to be able to compute max and min of the grid!
max and min
• Suppose you have a 2D grid of floating point numbers.
• min needs to be the lowest value in the grid, and max needs to be the highest.
22.5 25.5 28.3
21.5 15.5 29.0
29.1 18.2 19.9
max and min
min = 100000000.;
max = -100000000.;
for(i=0;i<NUMROWS;i++) {
for(j=0;j<NUMCOLS;j++) {
if (grid[i][j] < min) min = grid[i][j];
if (grid[i][j] > max) max = grid[i][j];
}
}
22.5 25.5 28.3
21.5 15.5 29.0
29.1 18.2 19.9
max and minmin = 100000000.;max = -100000000.;
for(i=0;i<NUMROWS;i++) {for(j=0;j<NUMCOLS;j++) {
if (grid[i][j] < min) min = grid[i][j];if (grid[i][j] > max) max = grid[i][j];
}}
• Set min to a very high number and max to a very low number.
22.5 25.5 28.3
21.5 15.5 29.0
29.1 18.2 19.9
max and min
min = 100000000.;
max = -100000000.;
for(i=0;i<NUMROWS;i++) {
for(j=0;j<NUMCOLS;j++) {
if (grid[i][j] < min) min = grid[i][j];
if (grid[i][j] > max) max = grid[i][j;
}
}
• For every element of the 2D array…
22.5 25.5 28.3
21.5 15.5 29.0
29.1 18.2 19.9
max and min
min = 100000000.;
max = -100000000.;
for(i=0;i<NUMROWS;i++) {
for(j=0;j<NUMCOLS;j++) {
if (grid[i][j] < min) min = grid[i][j];
if (grid[i][j] > max) max = grid[i][j];
}
}
• If grid[i][j] < min, min=grid[i][j] !
22.5 25.5 28.3
21.5 15.5 29.0
29.1 18.2 19.9
max and min
min = 100000000.;
max = -100000000.;
for(i=0;i<NUMROWS;i++) {
for(j=0;j<NUMCOLS;j++) {
if (grid[i][j] < min) min = grid[i][j];
if (grid[i][j] > max) max = grid[i][j];
}
}
• If grid[i][j] > max, max=grid[i][j] !
22.5 25.5 28.3
21.5 15.5 29.0
29.1 18.2 19.9
But Happily…
• mincon = (float) (( (int)(min/conint) + 1) * conint)• maxcon = (float) (( (int)(max/conint) ) * conint)
You can work out the math and see that this works! (maxcon is easier than mincon)
Grading Assignment 15
• Properly determine mincon and maxcon from a grid of data, prompting the user for the contour interval.
Grading Assignment 15
• Correctly determining most of the things you need to draw contours, but not getting good output: