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This Kodak Gray Scale, patch # 7 is an image of an 18% gray. The theory -- an object that reflects 18% of the ambient light is the middle of the photographic scale. If the camera exposure is spot-on and the film is developed to specification, the resulting negative (image of this object) will have a transmission density of 0.75 plus base fog density (about 0.10). If this negative is printed via spot-on exposure, and the photo paper is developed to specification, the image of this object will have a reflection density of 0.75. In other word, the object reflects 18% of the ambient light, the negative transmits 18% of the light that traverses the negative. The resulting print of this object reflects 18% of the ambient light. The 18% target shade is the same shade (brightness) original, negative, final display. The 0.75 density value.

Photo scientists use logarithmic notation base 10. Thus 10 elevated to the 0.75 power = 5.494. This would be the filter factor for a ND filter that corresponds to 18% transmission. The reciprocal 1/5.49 = 0.18. This decimal fraction expressed as a percent is 0.178 (0.18 rounded) X 100 = 18% (the reflection density).

Nobody said this stuff was easy! The 18% target has a reflection density of 18%. The RGB values for a monitor with a gamma of 2.2 (typical) reads 116R 116G 116B. Gamma is a measure of the steepness of a graph that measures contrast. Graph the readings of the gray scale and measure the angle of the upward sweep of the graph. We find this angle and then then using trigonometry find the TAN of this angle. This value is a measure of "contrast"

This Kodak Gray Scale, patch # 7 is an image of an 18% gray. The theory -- an object that reflects 18% of the ambient light is the middle of the photographic scale. If the camera exposure is spot-on and the film is developed to specification, the resulting negative (image of this object) will have a transmission density of 0.75 plus base fog density (about 0.10). If this negative is printed via spot-on exposure, and the photo paper is developed to specification, the image of this object will have a reflection density of 0.75. In other word, the object reflects 18% of the ambient light, the negative hold back 18% of the light that traverses the negative, the negative passes 82%. The resulting print of this object reflects 18% of the ambient light. The 18% target shade is the same original, negative, final display. The 0.75 density value.

Photo scientists use logarithmic notation base 10. Thus 10 elevated to the 0.75 power = 5.494. This would be the filter factor for a ND filter that corresponds to 18% transmission. The reciprocal 1/5.49 = 0.18. This decimal fraction expressed as a percent is 0.178 (0.18 rounded) X 100 = 18% (the reflection density).

Nobody said this stuff was easy! The 18% target has a reflection density of 18%. The RGB values for a monitor with a gamma of 2.2 (typical) reads 116R 116G 116B. Gamma is a measure of the steepness of a graph that measures contrast. Graph the readings of the gray scale and measure the angle of the upward sweep of the graph. We find this angle and then then using trigonometry find the TAN of this angle. This value is a measure of "contrast"

This Kodak Gray Scale, patch # 7 is an image of an 18% gray. The theory -- an object that reflects 18% of the ambient light is the middle of the photographic scale. If the camera exposure is spot-on and the film is developed to specification, the resulting negative (image of this object) will have a transmission density of 0.75 plus base fog density (about 0.10). If this negative is printed via spot-on exposure, and the photo paper is developed to specification, the image of this object will have a reflection density of 0.75. In other word, the object reflects 18% of the ambient light, the negative transmits 18% of the light that traverses the negative. The resulting print of this object reflects 18% of the ambient light. The 18% target shade is the same shade (brightness) original, negative, final display. The 0.75 density value.

Photo scientists use logarithmic notation base 10. Thus 10 elevated to the 0.75 power = 5.494. This would be the filter factor for a ND filter that corresponds to 18% transmission. The receptacle 1/5.49 = 0.18. This decimal fraction expressed as a percent is 0.178 (0.18 rounded) X 100 = 18% (the reflection density).

Nobody said this stuff was easy! The 18% target has a reflection density of 18%. The RGB values for a monitor with a gamma of 2.2 (typical) reads 116R 116G 116B. Gamma is a measure of the steepness of a graph that measures contrast. Graph the readings of the gray scale and measure the angle of the upward sweep of the graph. We find this angle and then then using trigonometry find the TAN of this angle. This value is a measure of "contrast"

This Kodak Gray Scale, patch # 7 is an image of an 18% gray. The theory -- an object that reflects 18% of the ambient light is the middle of the photographic scale. If the camera exposure is spot-on and the film is developed to specification, the resulting negative (image of this object) will have a transmission density of 0.75 plus base fog density (about 0.10). If this negative is printed via spot-on exposure, and the photo paper is developed to specification, the image of this object will have a reflection density of 0.75. In other word, the object reflects 18% of the ambient light, the negative transmits 18% of the light that traverses the negative. The resulting print of this object reflects 18% of the ambient light. The 18% target shade is the same shade (brightness) original, negative, final display. The 0.75 density value.

Photo scientists use logarithmic notation base 10. Thus 10 elevated to the 0.75 power = 5.494. This would be the filter factor for a ND filter that corresponds to 18% transmission. The reciprocal 1/5.49 = 0.18. This decimal fraction expressed as a percent is 0.178 (0.18 rounded) X 100 = 18% (the reflection density).

Nobody said this stuff was easy! The 18% target has a reflection density of 18%. The RGB values for a monitor with a gamma of 2.2 (typical) reads 116R 116G 116B. Gamma is a measure of the steepness of a graph that measures contrast. Graph the readings of the gray scale and measure the angle of the upward sweep of the graph. We find this angle and then then using trigonometry find the TAN of this angle. This value is a measure of "contrast"

This Kodak Gray Scale, patch # 7 is an image of an 18% gray. The theory -- an object that reflects 18% of the ambient light is the middle of the photographic scale. If the camera exposure is spot-on and the film is developed to specification, the resulting negative (image of this object) will have a transmission density of 0.75 plus base fog density (about 0.10). If this negative is printed via spot-on exposure, and the photo paper is developed to specification, the image of this object will have a reflection density of 0.75. In other word, the object reflects 18% of the ambient light, the negative transmits 18% of the light that traverses the negative. The resulting print of this object reflects 18% of the ambient light. The 18% target shade is the same shade (brightness) original, negative, final display. The 0.75 density value.

Photo scientists use logarithmic notation base 10. Thus 10 elevated to the 0.75 power = 5.494. This would be the filter factor for a ND filter that corresponds to 18% transmission. The receptacle 1/5.49 = 0.18. This decimal fraction expressed as a percent is 0.178 (0.18 rounded) X 100 = 18% (the reflection density).

Nobody said this stuff was easy! The 18% target has a reflection density of 18%. The RGB values for a monitor with a gamma of 2.2 (typical) reads 116R 116G 116B.

This Kodak Gray Scale, patch # 7 is an image of an 18% gray. The theory -- an object that reflects 18% of the ambient light is the middle of the photographic scale. If the camera exposure is spot-on and the film is developed to specification, the resulting negative (image of this object) will have a transmission density of 0.75 plus base fog density (about 0.10). If this negative is printed via spot-on exposure, and the photo paper is developed to specification, the image of this object will have a reflection density of 0.75. In other word, the object reflects 18% of the ambient light, the negative transmits 18% of the light that traverses the negative. The resulting print of this object reflects 18% of the ambient light. The 18% target shade is the same shade (brightness) original, negative, final display. The 0.75 density value.

Photo scientists use logarithmic notation base 10. Thus 10 elevated to the 0.75 power = 5.494. This would be the filter factor for a ND filter that corresponds to 18% transmission. The receptacle 1/5.49 = 0.18. This decimal fraction expressed as a percent is 0.178 (0.18 rounded) X 100 = 18% (the reflection density).

Nobody said this stuff was easy! The 18% target has a reflection density of 18%. The RGB values for a monitor with a gamma of 2.2 (typical) reads 116R 116G 116B. Gamma is a measure of the steepness of a graph that measures contrast. Graph the readings of the gray scale and measure the angle of the upward sweep of the graph. We find this angle and then then using trigonometry find the TAN of this angle. This value is a measure of "contrast"