Gets the scale factor of the font used to draw this chunk.

Namespace:  BitMiracle.Docotic.Pdf
Assembly:  BitMiracle.Docotic.Pdf (in BitMiracle.Docotic.Pdf.dll)

## Syntax

C#
`public double FontSize { get; }`
VB
```Public ReadOnly Property FontSize As Double
Get```

#### Property Value

Type: Double
The scale factor of the font used to draw this chunk.

## Remarks

A font defines the glyphs for one standard size. This standard is arranged so that the nominal height of tightly spaced lines of text is 1 unit. In the default user coordinate system, this means the standard glyph size is 1 point (or 1 ? 72 inch). The standard-size font must then be scaled to be usable.

This property retrieves the scale factor used to draw the text chunk. Usually, the FontSize property retrieves values like 10, 12, 14. But the FontSize might have an arbitrary value like 1, 1000, -1, -10. It's just a scale factor.

The final height of the text chunk depends on both FontSize and TransformationMatrix properties.

For a non-transformed text the TransformationMatrix is the identity matrix (the TransformationMatrix.IsIdentity() property is equal to true). In such a case the absolute text height is equal to Math.Abs(FontSize).

For a scaled text the TransformationMatrix looks like { M11; 0; 0; M22; 0; 0 }. In such a case the absolute text height is equal to Math.Abs(TransformationMatrix.M22 * FontSize).

For an arbitrarily transformed text the TransformationMatrix looks like { M11; M12; M21; M22; X; Y }. In such a case the absolute text height is equal to (Math.Abs(FontSize) * Math.Sqrt(Math.Pow(TransformationMatrix.M21, 2) + Math.Pow(TransformationMatrix.M22, 2))).

Let's try to understand the nature of this formula. At first, the matrix { FontSize, 0, 0, FontSize, 0, 0 } is multiplied by the current TransformationMatrix (current matrix being the right multiplier). The resulting matrix is { M11 * FontSize; M12 * FontSize; M21 * FontSize; M22 * FontSize; X; Y }. Then we calculate the text height in the new transformed space. We map points (0; 0) and (0; 1) from default space to the transformed space. The result is (X; Y) and (M21 * FontSize + X; M22 * FontSize + Y), respectively. And finally we calculate the length of this vector, that is exactly the absolute text height in the transformed space: Math.Sqrt(Math.Pow(M21 * FontSize, 2) + Math.Pow(M22 * FontSize, 2)) = Math.Abs(FontSize) * Math.Sqrt(Math.Pow(M21, 2) + Math.Pow(M22, 2))

Below is an example of rotated text for which the FontSize is equal to 1.