build a circle using the latitude and longitude values

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Angela Marino
Angela Marino 2020 年 7 月 2 日
編集済み: Adam Danz 2020 年 7 月 6 日
Hi, I need to build a circle using the latitude and longitude values ​​as the center of the circle.
busStop_latit; %array (28,1)
busStop_long; %array (28,1)
%latit and long with which I want to use to build the circle
I need to define "true" if new latitude and longitude values ​​are internal to the constructed circle, "false" if these new values ​​are external to the circle. How is it possible to build the circle and write these two conditions in matlab?
latitude; %matrix (3600,4)
longitude; %matrix (3600,4)
%i want to use only the 1st column oh the two matrix
pos_busStop=ones(28,1);
distance; %matrix (3600,4) and I want to use only the 1st column for this mtrix too
idx=zeros(3600,1);
for i=1:3600
if lat(i,1)<lat_new(1,:) && lon(i,1)<long_new(1,:)
idx(:,1)=1
end
pos_busStop(:,1)==distanza(idx,1);
end
  3 件のコメント
Angela Marino
Angela Marino 2020 年 7 月 2 日
Yes, but i don't know how to write the code
Adam Danz
Adam Danz 2020 年 7 月 2 日
On second thought, pdist2() with 3600 coordinates may overwhelm that function. It would be easiest just to use the euclidean distance equation directly. I'll add a demo in the answers section that you can apply to your data.

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Adam Danz
Adam Danz 2020 年 7 月 2 日
Here's a demo that you can apply to your data.
% Define the center of the circle
target = [lon,lat]; % where lon and lat are single, "scalar" values
% Define the radius of the circle (distance from target to furthest accepted point)
r = 12.2;
% Compute the distance of each (lon,lat) coordinate from target
% lonVec is a vector of longitudinal coordinates
% latVec is a vector of latitudinal coordinates
% hypot(A,B) is a Matlab function that computes the Euclidean distance between (A,B)
dist = hypot(lonVec-target(1), latVec-target(2));
% Detect which coordinates are within range
withinRange = dist <= r; % withinRange is a logical vector the same size as lonVec | latVec
  6 件のコメント
Angela Marino
Angela Marino 2020 年 7 月 3 日
pos_fermateR2=ones(28,1);
% Define the center of the circle
target = [long_ferm(:,1),lat_ferm(:,1)]; % where lon and lat are single, "scalar" values
% Define the radius of the circle (distance from target to furthest accepted point)
r = 12.2*ones(28,1);
% Compute the distance of each (lon,lat) coordinate from target
% lonVec is a vector of longitudinal coordinates
% latVec is a vector of latitudinal coordinates
% hypot(A,B) is a Matlab function that computes the Euclidean distance between (A,B)
for i=1:3600
dist = hypot(abs(lon(i,1)-target(:,1)), abs(lat(i,1)-target(:,2)));
% Detect which coordinates are within range
% withinRange is a logical vector the same size as lonVec | latVec
withinRange = dist <= r;
if veloc(i,1)==0 && veloc(i-1,1)~=0
pos_fermateR2(:,1)==distanza(withinRange,1);
end
end
Adam Danz
Adam Danz 2020 年 7 月 6 日
編集済み: Adam Danz 2020 年 7 月 6 日
Try using pdist2. Your lat and lon values contain a lot of NaNs which will propagate through to the final output.
target = [long_ferm(:,1),lat_ferm(:,1)];
dist = pdist2(target, [lon(:,1),lat(:,1)]);
inCirlce = dist <= 12.2;
inCircle is a matrix of logical values.
size(inCirlce)
ans =
28 3600
where inCircle(i,j) tells you whether [lon(j,1),lat(j,1)] is within range of the target target(i,:). 1 means in or on cirlce and 0 means outside or NaN value was present.

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