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- 1970-1-1
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发表于 2023-3-26 23:50:32
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- % 读取数据
- nc_path = '202007.nc'; % 数据路径
- ncid = netcdf.open(nc_path, 'nowrite');
- lon = netcdf.getVar(ncid, netcdf.inqVarID(ncid, 'longitude'));
- lat = netcdf.getVar(ncid, netcdf.inqVarID(ncid, 'latitude'));
- level = netcdf.getVar(ncid, netcdf.inqVarID(ncid, 'level'));
- time = netcdf.getVar(ncid, netcdf.inqVarID(ncid, 'time'));
- u = netcdf.getVar(ncid, netcdf.inqVarID(ncid, 'u'));
- v = netcdf.getVar(ncid, netcdf.inqVarID(ncid, 'v'));
- vo = netcdf.getVar(ncid, netcdf.inqVarID(ncid, 'vo'));
- netcdf.close(ncid);
- % 计算涡度方程各项
- [lon1, lat1] = meshgrid(lon, lat);
- V1_all = zeros(length(lon), length(lat));
- V4_all = zeros(length(lon), length(lat));
- for x = 1:length(lon1); % 经度范围
- for y = 1:length(lat1); % 纬度范围
- z0 = 1; % 垂直层数
- t0 = 1; % 时间步长
- fi = lat(y) * pi / 180; % 纬度转弧度
- beta = 2 * cos(fi) * 7.29e-5; % beta参数
-
- % 计算dvo_dx和dvo_dy
- dx = (lon(2) - lon(1)) * pi / 180; % 计算经度间隔
- dy = (lat(2) - lat(1)) * pi / 180; % 计算纬度间隔
- dvo_dx = double(diff(vo(x, :, z0, t0), 1, 2)) ./ double(dx) * 6371000 * cos(fi); % x方向上的vo梯度
- dvo_dy = double(diff(vo(:, y, z0, t0), 1, 1)) ./ double(dy) * 6371000; % y方向上的vo梯度
- % 补充缺失的最后一个元素
- dvo_dx = [dvo_dx, dvo_dx(end)];
- dvo_dy = [dvo_dy; dvo_dy(end)];
- V1 = -double(u(x, y, z0, t0)) .* dvo_dx - double(v(x, y, z0, t0)) .* dvo_dy;
- V4 = double(v(x, y, z0, t0)) .* beta;
- V1_all(x, y) = V1; % 将计算结果存储到数组中
- V4_all(x, y) = V4;
- end
- end
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盲写的,我也不知道能不能解决你的问题,如果有报错就更好了 |
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