Recent content by jebez

I think it's like audio on a PC: sub-pixel value (0 to 255) goes to a DAC (digital to analog converter) then amplified, multiplied (brightness) by a transistor. I googled a lot without finding, confirmation...

I care not if sub-pixel value & output power is linear or not, I'm interested by the relationship between the 3 sub-pixels, back to my 1st post: Static image on a OLED display at brightness 0%, a pixel has 3 sub-pixels emitting r, g & b watt. I increase the brightness: now they emit rx, gx & bx...

Well e.g. a red sub-pixel has the value 142, at brightness 0% it emits x watt, I increase the brightness to 24%: it emits 24*x or 24*a+x watt? a is constant.

Value of a sub-pixel: on 8 bits, 1 octet https://en.wikipedia.org/wiki/Color_depth#True_color_(24-bit) , 2^8=256 values possibles, 256^3=16 777 216 colors possible (3 sub-pixels). Brightness: Arch Linux KDE, panel, Screen Brightness from 0 to 100%. Aggravating_Bear9268 in the reddit link...

I can't edit my 1st post, clearer: Increasing the brightness of an emissive display, how a pixel? Emissive display: OLED, CRT etc, not LCD. Are the values of the 3 sub-pixels added or multiplied by a same value, or other? Is it R+x, G+x & B+x or nothing if a sub-pixel is off, R*x, G*x & B*x or...

omg my 1st post isn't clear?

Huh? The brightness of my ViewSonic VP-16 OLED monitor can be increased from 0 to 100%, by my PC running Arch Linux KDE...

Increasing the brightness of an OLED pixel (display in general), the 3 sub-pixels are added or multiplied? Is it R+x, G+x & B+x or R*x, G*x & B*x ? (Red Green Blue). [Posted edited by a Mentor]

Earth's elliptic orbit : assume(ε>0,ε<1); plot2d(2456084229048821270250000000/489238659320312440009*subst(ε=0.0167086,integrate(1/(1+ε*cos(θ))^2,θ)),[θ,-π,π]);

Erratum : https://en.wikipedia.org/wiki/Kepler_problem instead of https://en.wikipedia.org/wiki/Kepler's_equation in my #21 post . In Maxima : assume(ε>0,ε<1); integrate(1/(1+ε*cos(θ))^2,θ); assume(ε>0,ε<1,θ>-π,θ<π); solve(t=integrate(1/(1+ε*cos(θ))^2,θ),θ);

https://en.wikipedia.org/wiki/Kepler's_equation just gives at the end the Kepler's 1st law of my #17 post in general ( ##-\frac{k}{m}##instead of ##GM## ) . The problem is solved , we've r(θ) & t(θ) , the summary with " the conversation ultimately ends without a solution " needs to be edited ...

In : var('θ,ε,t') assume(0<ε<1) assume(-π<θ<π) from sage.symbolic.integration.integral import indefinite_integral latex(solve(t==indefinite_integral(1/(1+ε*cos(θ))^2,θ),θ)) Out : ##\left[\sin\left(θ\right) = \frac{\sqrt{-ε^{2} + 1} {\left(ε \cos\left(θ\right) + ε\right)} - \sqrt{-t^{2} ε^{8} +...

I can't edit my previous post ... There're (cos(θ)+1) as denominators , so we must precise : -π<θ<π

I'm back , sorry for long time . In short : ##L=r^2ω=constant## ( Kepler's 2nd law ) ##r=\frac{L^2}{GM(1+εcos(θ))}## ( Kepler's 1st law ) ⇒##\frac{1}{ω}=\frac{dt}{dθ}=\frac{L^3}{(GM(1+εcos(θ)))^2}## ⇔##t(θ)=\frac{L^3}{(GM)^2}∫\frac{1}{(1+εcos(θ))^2}dθ## ##∫\frac{1}{(1+εcos(θ))^2}dθ## ...

t(θ)=$$\frac{a^2 \left(\frac{b (x \sin )}{\left(b^2-1\right) (b (x \cos )+1)}-\frac{2 (b-1) (x \tan ) (x \coth )}{2 \left(b^2-1\right)^2}+d\right)}{c}$$ ( a=p , b=ε , c=c , d=t0 ) given by Mathematica ( I've 15 days trial ) , not WolframAlpha . There's also the differential equation...