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Contouring in C ATS 315
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How Contours Really Work Contours may LOOK like curves, but they are really just straight line segments.
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How Contouring Works:
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So What We Are Really Worried About…
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The northwest corner is grid[i][j]. grid[i][j]
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The northwest corner is grid[i][j]. grid[i][j]grid[i][j+1] grid[i+1][j]grid[i+1][j+1]
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The northwest corner is grid[i][j]. grid[i][j]grid[i][j+1] grid[i+1][j]grid[i+1][j+1]
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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]
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For simplicity, rename the values at the corners: nwne swse
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Count the number of corners with values greater than the contour line. nwne swse
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CornersGreaterThanContour can be 0, 1, 2, 3, or 4. nwne swse
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if (CornersGreaterThanContour==0) do nothing nwne swse
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if (CornersGreaterThanContour==4) do nothing nwne swse
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if (CornersGreaterThanContour==1)… there are four possibilities: nwne swse
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if (CornersGreaterThanContour==1)… there are four possibilities: nwne swse Possibility 1: Only the northwest corner is greater than the value of the contour.
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if (CornersGreaterThanContour==1)… there are four possibilities: nwne swse Possibility 2: Only the northeast corner is greater than the value of the contour.
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if (CornersGreaterThanContour==1)… there are four possibilities: nwne swse Possibility 3: Only the southeast corner is greater than the value of the contour.
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if (CornersGreaterThanContour==1)… there are four possibilities: nwne swse Possibility 4: Only the southwest corner is greater than the value of the contour.
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if (CornersGreaterThanContour==2)… there are three possibilities: nwne swse
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if (CornersGreaterThanContour==2)… there are three possibilities: nwne swse 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.
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if (CornersGreaterThanContour==2)… there are three possibilities: nwne swse 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.
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if (CornersGreaterThanContour==2)… there are three possibilities: nwne swse 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.
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if (CornersGreaterThanContour==3)… there are four possibilities: nwne swse
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if (CornersGreaterThanContour==3)… there are four possibilities: nwne swse Possibility 1: Only the northwest corner is less than the value of the contour.
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if (CornersGreaterThanContour==3)… there are four possibilities: nwne swse Possibility 2: Only the northeast corner is less than the value of the contour.
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if (CornersGreaterThanContour==3)… there are four possibilities: nwne swse Possibility 3: Only the southeast corner is less than the value of the contour.
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if (CornersGreaterThanContour==3)… there are four possibilities: nwne swse Possibility 4: Only the southwest corner is less than the value of the contour.
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What will this program look like?
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/* Compute CornersGreaterThanContour */ if (CornersGreaterThanContour == 0) { } if (CornersGreaterThanContour == 1) { } if (CornersGreaterThanContour == 2) { } if (CornersGreaterThanContour == 3) { } if (CornersGreaterThanContour == 4) { }
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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 */ }
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What will this program look like? if (CornersGreaterThanContour == 1) { } if (CornersGreaterThanContour == 2) { } if (CornersGreaterThanContour == 3) { }
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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) { }
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What will this program look like? if (CornersGreaterThanContour == 2) { } if (CornersGreaterThanContour == 3) { }
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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) { }
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What will this program look like? if (CornersGreaterThanContour == 3) { }
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What will this program look like? if (CornersGreaterThanContour == 3) { if( nw < contour) { } if( ne < contour) { } if( se < contour) { } if( sw < contour) { } }
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Deciding how to draw the contour nwne swse 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.
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Deciding how to draw the contour nwne swse What determines the location of the starting point?
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Deciding how to draw the contour nwne swse What determines the location of the starting point? INTERPOLATION!
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Deciding how to draw the contour nwne swse Suppose we are drawing the 1000 mb contour. nw = 999.9 ne = 1022.1
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Deciding how to draw the contour nwne swse Suppose we are drawing the 1000 mb contour. nw = 982.2 ne = 1001.1
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Deciding how to draw the contour nwne swse startlat = ??? startlon = ???
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Deciding how to draw the contour nwne swse 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
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Deciding how to draw the contour nwne swse interp(gridlatitude[i][j], &startlat, gridlatitude[i][j+1], nw, contour, ne); interp(gridlongitude[i][j], &startlon, gridlongitude[i][j+1], nw, contour, ne);
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Deciding how to draw the contour nwne swse How about the endlat and endlon?
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Deciding how to draw the contour nwne swse 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);
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Deciding how to draw the contour nwne swse Once you have (startlat, startlon) and (endlat, endlon): 1.tranform 2.clip 3.gline
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How this produces contours
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i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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How this produces contours i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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How this produces contours i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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How this produces contours i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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How this produces contours i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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How this produces contours i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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How this produces contours i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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How this produces contours i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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for(contour=mincontour;contour<=maxcontour;contour=contour+conint) i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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for(contour=mincontour;contour<=maxcontour;contour=contour+conint) i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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for(contour=mincontour;contour<=maxcontour;contour=contour+conint) i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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for(contour=mincontour;contour<=maxcontour;contour=contour+conint) i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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for(contour=mincontour;contour<=maxcontour;contour=contour+conint) i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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for(contour=mincontour;contour<=maxcontour;contour=contour+conint) i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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for(contour=mincontour;contour<=maxcontour;contour=contour+conint) i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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for(contour=mincontour;contour<=maxcontour;contour=contour+conint) i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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for(contour=mincontour;contour<=maxcontour;contour=contour+conint) i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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for(contour=mincontour;contour<=maxcontour;contour=contour+conint) i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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for(contour=mincontour;contour<=maxcontour;contour=contour+conint) i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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for(contour=mincontour;contour<=maxcontour;contour=contour+conint) i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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for(contour=mincontour;contour<=maxcontour;contour=contour+conint) i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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for(contour=mincontour;contour<=maxcontour;contour=contour+conint) i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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for(contour=mincontour;contour<=maxcontour;contour=contour+conint) i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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for(contour=mincontour;contour<=maxcontour;contour=contour+conint) i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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for(contour=mincontour;contour<=maxcontour;contour=contour+conint) i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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for(contour=mincontour;contour<=maxcontour;contour=contour+conint) i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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for(contour=mincontour;contour<=maxcontour;contour=contour+conint) i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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for(contour=mincontour;contour<=maxcontour;contour=contour+conint) i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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for(contour=mincontour;contour<=maxcontour;contour=contour+conint) i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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for(contour=mincontour;contour<=maxcontour;contour=contour+conint) i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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for(contour=mincontour;contour<=maxcontour;contour=contour+conint) i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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for(contour=mincontour;contour<=maxcontour;contour=contour+conint) i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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for(contour=mincontour;contour<=maxcontour;contour=contour+conint) i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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for(contour=mincontour;contour<=maxcontour;contour=contour+conint) i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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for(contour=mincontour;contour<=maxcontour;contour=contour+conint) i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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for(contour=mincontour;contour<=maxcontour;contour=contour+conint) i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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for(contour=mincontour;contour<=maxcontour;contour=contour+conint) i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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for(contour=mincontour;contour<=maxcontour;contour=contour+conint) i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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for(contour=mincontour;contour<=maxcontour;contour=contour+conint) i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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for(contour=mincontour;contour<=maxcontour;contour=contour+conint) i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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for(contour=mincontour;contour<=maxcontour;contour=contour+conint) i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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for(contour=mincontour;contour<=maxcontour;contour=contour+conint) i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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for(contour=mincontour;contour<=maxcontour;contour=contour+conint) i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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for(contour=mincontour;contour<=maxcontour;contour=contour+conint) i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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for(contour=mincontour;contour<=maxcontour;contour=contour+conint) i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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for(contour=mincontour;contour<=maxcontour;contour=contour+conint) i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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for(contour=mincontour;contour<=maxcontour;contour=contour+conint) i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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for(contour=mincontour;contour<=maxcontour;contour=contour+conint) i=0 i=1 i=2 i=3 j=0j=1j=2j=3
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Great, I can picture what this looks like, but how do I do it?
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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….)
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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) { } }
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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…)
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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.)
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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.)
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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.
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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
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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!
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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. 22.525.528.3 21.515.529.0 29.118.219.9
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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]; } 22.525.528.3 21.515.529.0 29.118.219.9
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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]; } Set min to a very high number and max to a very low number. 22.525.528.3 21.515.529.0 29.118.219.9
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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… 22.525.528.3 21.515.529.0 29.118.219.9
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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] ! 22.525.528.3 21.515.529.0 29.118.219.9
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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] ! 22.525.528.3 21.515.529.0 29.118.219.9
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But Sadly… min mincon max maxcon
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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)
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Grading Assignment 15 Properly determine mincon and maxcon from a grid of data, prompting the user for the contour interval.
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Grading Assignment 15 Correctly determining most of the things you need to draw contours, but not getting good output:
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Grading Assignment 15 Successfully contouring temperature fields:
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Grading Assignment 15 Successfully contouring any grid of data chosen by the user:
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Suggested “Impressive” Things Variable domains Variable colors for the contours Nice labels for the plot (NOT contour labels!)
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