now divide all the exposure value triangles into equal parts and they will/ should hit the same bulls eye 😎
March 25, 2026
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What?!! Please take the time to post an illustrative diagram or two.
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[s]For an exposure triangle, is the point C always in the middle of the inclined triangular plane such that x=y=z ?[/s]
entered in error, Arvo, oops
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typo? x+y+x?
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Typo of course, sorry.
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No problem!
reading Wikipedia about Euclidian planes - heavy going - even from the bottom up at my age ...
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<deleted per Arvo comment>
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Please, this goes off of even current topic. Assume just anything 'normally exposed', looking good on screen, OK?
As we here attempt to (re)invent some mathematics of exposure triangle, then the most important thing is to establish relation between input parameters, keeping output (brightness) unchanged.Now back to your first triangle, I thought about it a bit more.
Apparently such triangle is usable for constant brightness (or constant overall system gain, whatever definition suits you better). In graphical form - if you read any point "coordinates" as I draw on next image (blue lines, perpendicular to corresponding sides), then they give three parameter values, giving same brightness for the same scene (1).If you need to change brightness (gain) or scene illumination), then you need to shift at least one of the scales (like shift ISO to left or right). In resulting triangle again every point corresponds to same result. (Of course depth of field, movement effects and noise will be different - but brightness remains constant).
(1) Actually not on current image - you need exactly similar logarithmic scales on every side; this means that they must cover exactly same amount of steps. On your triangle apertures spans over 7 steps, other parameters over 6 - this way relations do not hold.
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<dupe deleted by author>
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good point - not all published illustrations are perfect in that regard ...
... in other words a published "stop" is not always the same geometric length on each side - especially if the sides don't have an equal number of marks.
Getting clearer by the minute, thanks again!
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So, in this one, the two points blue and green are the same brightness as all the other points inside the triangle, ignoring for now the different stop scaling of the sides?
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Yes.
On your image (approximately), blue gives f/8, 1/125, ISO 400 and green f/4 (+2 stops), 1/250 (-1), ISO 200 (-1) - result is same. -
If green and blue dots represent the same brightness shouldn't they be in the same position on the diagram?
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The combination of these 3 settings for each of these 2 dots gives the same brightness.
e.g. if you start with one dot anywhere within the triangle and change the shutter speed by +1 stop to make the image 1 stop brighter , and at the same time change the aperture by -1 stop you'll get the same brightness, but the two dots will have different positions. So you'll end up with 2 dots in different places, but with the same brightness.
In fact all dots within that triangle will have the same defined brightness
Please note I'm (also) ignoring the "typo" error (already pointed out), that the aperture side on this particular triangle picture has 7 stops, but should really only show 6 stops like the other 2 sides of the triangle)here's an idea,...
An interesting addition to the diagram would be the the possibility to "slide the scale" on say one side of the triangle to create another different desired brightness of all dots within this "new" trianglee.g. slide the ISO scale, by one position (1 stop) to the left, so that it now starts with ISO 200 and goes up to ISO 6400.
Now, all points within this "new" triangle would be "twice" as bright as before doing the slide (applies to the RAW data)
You could also slide the triangle's scales for shutter speed or aperture, to create some new desired defined brightness for the triangle -
Seems like the Triangle relates to Exposure Value a la Wikipedia ...
... from which Ev = log2(A^2 / t) + log2(S/100) i.e. Ev = 2log2(A) - log2(t) + log2(ISO/100) where taking logs to base two relates to the "stops" presentation of the three parameters.
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Less triangular, I could draw a nomogram of three parallel lines ... 'A' at the top, 'Ev' in the middle and shutter speed 't' at the bottom. Then scale them such that a straight line drawn from a point on 'A' to a point on 't' would pass through the Ev for that combination on the middle line.
OK for someone like me who always shoots at 100 ISO, ho-ho.
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Hello Donald ....... ?
