class VectorAlgebra
{
public static Vector3 Subtract(Vector3 a, Vector3 b)
{
return new Vector3(a.x-b.x, a.y-b.y, a.z-b.z);
}
public static Vector3 Scale(float f, Vector3 a)
{
return new Vector3(f*a.x, f*a.y, f*a.z);
}
public static float Dot(Vector3 a, Vector3 b)
{
return (a.x*b.x)+(a.y*b.y)+(a.z*b.z);
}
public static Vector3 Projection(Vector3 a, Vector3 b)
{
return Scale(Dot(a, b)/Dot(b, b), b);
}
public static Vector3 Rejection(Vector3 a, Vector3 b)
{
return Subtract(a, Projection(a, b));
}
static void Main(string[] args)
{
var A=new Vector3(5, 5, 0);
var B=new Vector3(0, 10, 0);
var C=Rejection(A, B);
// C = { 5,0,0}, expected answer from math {5,5,0}-0.5*{0,10,0}
}
}
# How To: Display a map at the proper reference scale
## Summary
In order to conceptualize the size of features and distances on a map, we must establish and indicate a consistent relationship between size on the map and size in real life. This relationship is the map's reference scale. For instance, a map may use one inch to represent what is actually 24,000 inches on earth. The reference scale is expressed as a mathematical ratio, or fraction where the unit on the map is expressed as the number one (numerator). A scale of 1:24,000 would be read as "one to twenty-four thousand." A value such as 24,000 inches cannot be easily visualized, therefore maps usually convert the ratio to another unit of measurement. For example, a 1:24,000 map can also be expressed as 1 inch = 2000 feet.
Concepts for defining a reference scale for a map are cross-application, namely they apply to all mapping applications. When working with an ArcIMS image service, the output map is generated as an image, thus reference scale concepts must be to applied to the contents of the map image. When displaying map images in an application, rendering the map image at the proper scale depends on three variables: the resolution of the image and the resolution and display area of the clients monitor, and the map extent. The image size and monitor resolution are measured in pixels while a monitor's display area and the map extent is expressed using a measurable unit like inches or meters. In general, on a computer screen, reference scale equals:
(map units / pixel) * (pixels / inch of display area) * (screen units / map unit)
The number of map units per image pixel (map units / pixel) is commonly referred to a the Map Scale Factor and can be easily calculated. For example:
(maximum x value for map extent - minimum x value for map extent) / width of map image in pixels
It is somewhat more difficult to determine how many pixels are rendered per inch of display area (pixels / inch of display area) since the monitor display size and resolution must be known - usually referred to as ppi (pixels per inch). Unfortunately, client-side scripting (i.e. JavaScript, VBScript) cannot retrieve a client's viewable display area in a measured unit, therefore we must prompt the user to provide this information. We can, however, return the screen resolution in pixels, which will be used to calculate the ppi.
The number of screen units in one map unit (screen units / map unit) is a constant value depending on which units are being used. In most applications, the screen is measured using inches while the map unit is in meters or feet. For example, there are 39.3701 inches in 1 meter.
See the following example to calculate the new extent of a map image given a specific reference scale and monitor display width:
2) The Map Scale Factor can be used to calculate the width and height (in map units) of the new map extent. Multiply the Map Scale Factor by the map image width and height to return the new map extent width and height:
3) To calculate the new map extent, the difference between the initial and new map width and height must be determined. First the absolute width and height of the initial map image are calculated:
4) To change the extent parameters to reflect this change, the absolute difference for both width and height must be divided by 2, then added to the initial maximum x and y values and subtracted from the initial minimum x and y values:
New xmin = 187696.6978 - (200521.7028 / 2) = 228897.9717
New xmax = 200594.4110 + (200521.7028 / 2) = 342602.5978
New ymin = 282314.4923 - (106833.041 / 2) = 87435.8464
New ymax = 289186.0772 + (106833.041 / 2) = 300855.2624
The new map extent envelope can be defined as needed.
