Understanding Pre-Condenser lens

[Rox]

now here is a problem there, virtual arc size tells the lens placement, and the lens to arc placemnt determines virtual arc size... this is like iterative problem.

Iterative problem is when you need to give one value to someting and then work out all other values and see if the first value is fine, if it is not, then change it increasing or decreasing (it depends on the last time error direction) and try again, finally you get to the result close enough.

anyway I´ll try to work out the ideal placement including arc lengh as one parameter to be considered as well. Let´s work it out

[Rox]

This is the problem, there are all the variables drown.

V= virtual arc half lengh.
A= real arc half lengh.
D=half precondensor lens diameter.
F=focal of precondensor lens.
d=arc--precondensor lens distance.
t=virtual arc---precondensor lens dist.
f=rear fresnell focal lengh.
L=half lcd or half rear fresnell diagonal.



The problem is to work out d and t so we know where to place the precondensor lens. The following data is supossed to be known: A, D, F, f, L.

The equations are well known also;

1/F - 1/d =-1/t

A*t/d=V

now, if you like doing maths, it is a good problem. Any interested there?

[Sarna]

You never rest mate?
Your 2 equation three variables system has no finite solution, but if your draw and mathematical statements are correct (haven't checked them myself), you can squeeze them a little more and solve the system. I've done it as you seem unable to (just guessing, you haven't proven to be lazy though far ). It's noted down in a hurry while cooking so I cant really swear there's no error in it, though I'll work it again tonite to be safe...

It reduces to a quadratic equation ax2+bx+c=0
x=d
a=L+D
b=f(2A-2D-L)
c=f2(D-A)
A is your "real arc halflenght", f2 is square...

I'm sure anyone can solve this basic mathematics, and perhaps there's a more fancy solution but I'm not gonna waste hours to know if there is myself.

[Rox]

i´ll have a look to your maths right now, but first would like to tell you there are more equations implied than what i posted (for instance you can get the relation of V and D just doing it intersect with the light tringle.)

i think i get to a solution also, but am testing it before postine here.

[Rox]

there is something wrong on your math, There should be F somewhere (F= the focal lengh of precondenser lens). It is very important variable.

There are same equations as variables, this means there is only one solution for it. I think the easiest way findig the solution is the iterative way. But I think my solution is fine as well, just let me check something more before posting...

good try Rox

[Sarna]

Quoting myself... "if your draw and mathematical statements are correct...you can squeeze them a little more"
I simply solved your drawing, nothing more...

[Rox]

i don´t understand what you did but if i give you A, F, D, f, L values, would you tell me V, d, t values?

some people like doing crosswords instead those things , well i am not one of them

[Sarna]

I'm a busy man mate, only tried to help you as I've seen you arguing in every forum and it's neighbour lately, no offense intended or taken plz
Of course I'd give you the values, I got an A+ in advanced mathematics every course at the university, I made some money teaching mathematics in there last six years, and these are only high school problems... it would help if you include the supposed equations or at least draw all the variables.

[Rox]

ok, take your time (tomatelo con calma );

all the variables are drown, and the equations are those I wrote and there are few more you can work out from the picture;

the keys are;
1) virtual image is at rear fresnell's focal point
2) the precondensor diameter is that needed to intersect with the virtual arc outer light ray and the lcd corner.

You don´t need more information but the known parameters;

A=12mm (half of 24mm arc)
F=100mm precondensor lens focal
D=50mm precondenser Radius (half of 100 mm diameter)
f=330mm rear fresnell focal
L=190mm half 15" lcd.

the solution is real and single. please let me know V, d, t. Suerte

[Sarna]

You still didn't provide an equation or drawing that made use of the condensor focal length, so no way of including it in the results
I assume that the condensor Focal would be one of those I've put in your drawing, if it is tell me if it's the red or the black one, both have easy and quick numeric solution. If it's not please draw the condensor focal in the picture yourself. Perhaps it's obvious to you but I haven't read the whole thread, only your mssg asking for help...