Help solve video bitrate equation

Childs

Lifer
Jul 9, 2000
11,313
7
81
OK, so I havent had to simplify an equation in awhile, and I apparently can't remember how to do this sheit, so I need to cheat. How do you simply (s - (a * l))/l = v to get l? I end up canceling out l, which is what I'm trying to find.
 

cjr22

Member
Mar 21, 2003
65
0
0
Multiply both sides by I, add a*I to both sides and you get s=a*I + v*I.
Then divide by a+v and you get I= s/(a+v)
 

Childs

Lifer
Jul 9, 2000
11,313
7
81
So for l:

(s - (al))/l = v

l * (s - (al))/l = v * l

s - (a * l) = vl

(al) + (s - (al) = vl + (a*l)

s = vl + al

s = l(v+a)

1/(v+a) * s = l(v+a) * 1/(v+a)

s/(v+a) = l

For a:

l = s/(v+a)

1/s * l = v+a * 1/s

l/s = v + a

l/s -v = a

a = s/l - v

?
 

OCedHrt

Senior member
Oct 4, 2002
613
0
0
l = s / ( v + a )
lv + la = s
la = s - lv
a = s/l - v

or following what you were trying to do:

l = s / ( v + a )
1/s * l = s / ( v + a ) * 1/s
l/s = 1 / ( v + a )
l(v + a) = s
lv + la = s
la = s - lv
a = s/l - v



Originally posted by: Childs
So for l:

(s - (al))/l = v

l * (s - (al))/l = v * l

s - (a * l) = vl

(al) + (s - (al) = vl + (a*l)

s = vl + al

s = l(v+a)

1/(v+a) * s = l(v+a) * 1/(v+a)

s/(v+a) = l

For a:

l = s/(v+a)

1/s * l = v+a * 1/s

l/s = v + a

l/s -v = a

a = s/l - v

?

 

Childs

Lifer
Jul 9, 2000
11,313
7
81
Thanks everyone. I passed this equation around work and everyone came up with a different solution, especially when solving for the other variables. It starting to come back, but sheesh, 1 min project turned into an all day affair. Time to buy an Algebra book.