将我的 WGS 84 坐标转换为屏幕 x 和 y
我在创建一个简单的地图应用程序时遇到了一个小问题。我想要学习的是 WGS 84 GPS 坐标转换背后的数学。
这就是问题：
我有两个 WGS 84 坐标；纬度：53.7506 长：-2.3931 + 纬度：53.7638 长：-2.4212 我想将这些值转换为屏幕上的点 x = 'a' 值在 0 到 200 之间，y = 'a' 值在 0 到 200 之间
我已经说了一个值，因为该值将从纬度和经度数字导出，如下所示上面提到过。
有人可以帮忙吗？
我尝试使用公式 N = V/D 对数字进行标准化（N 是标准化值，V 是坐标值，D 是屏幕区域的最大值，在我的例子中是 200。这并不起作用生成的数字都在 199 - 200 范围内，但我现在已经迷失了这一点，所以我需要一些帮助，
我确信有一个计算正确的公式，但我似乎找不到它，
请帮忙。我需要今晚睡个好觉:)
谢谢
I am stuck on a little problem of creating a simple map app. what i am trying to learn is the mathematics behind the conversion of my WGS 84 GPS co-ordinates.
this is the problem:
I have two WGS 84 co-ordinates; lat:53.7506 Long:-2.3931 + lat:53.7638 Long:-2.4212
I want to convert these values onto points on my screen x = 'a' value between 0 and 200 and y = 'a' value between 0 and 200
i have said a value as this value will be derived from the lat and long figures as mentioned above.
can anyone help please?
i have tried normalising the figures using the formula N = V/D (N being the normailsed value with V being the co-ordinate value and D being the max value of the screen area which in my case is 200. this didnt work as all the figures generated was all in the 199 - 200 range but I have gotten lost with the thought process on this now so I need some help.
I am sure there is a formula for calculating the right but i cannot seem to find it
please help as I need to sleep well tonight :)
thanks
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这是 x 的简单线性映射，以及带有扭曲的 y 的线性映射：
x) 因为下限是左侧，您可以简单地使用
y)，因为只要 0 在中间，您就需要减去/相应添加：
It is a simple linear mapping for x, and one for y with a twist:
x) as the lower limit is the left hand side, you can simply use
y) as long 0 is in the middle, you'll need to substract/add accordingly: