Involution or square root in an expression?

[pvd]

Hi all,

 

Can anybody tell me how I can put an involution or a square root in an expression?

 

Thanks in advance!

[JD Mather]

Can you attach what you have so far and also indicate here the exact (including units) expression you are trying to use?

[pvd]

It's an expression with dimensions i used before (d10 and d11, both in mm) but I don't think that's a problem right? The result should be in mm as I'm talking about a length:

 

(1-d10² ul/d11² ul)^0.5 * d11

[JD Mather]

Your expression doesn't make sense. Not handling units correctly. Can you zip and attach the ipt here?

[shift1313]

unless d10^2 is less than 1 you have the square root of a negative number which is an invalid expression.

 

what is this equation giving you? An involution is a function that is its own inverse. When you talk about inverse and square functions you need a specific domain. what are you working with here?

[JD Mather]

(d10 in mm) d10² ul

 

Lets look at a small part of the equation.

You can't simply assign units.

If d10=1mm

then

d10^2 = 1mm^2

therefore

1mm^2 ul does not make sense.

 

You would need to multiply this by a factor to remove the mm^2 units. (1/1mm^2)

[pvd]

Ok guys, back again:

 

I attached the ipt hoping to be more clear now. It's about an ellipse: large axis 200 (d0), small axis 60 (d1).

I would like to find the focal points on this ellipse by putting an expression for d2

 

The expression should be: (forgive me if I'm wrong, it's the theory that counts, right?)

 

(1 - 60^2/200^2)^0.5 * 200

or

(1 - d1^2/d0^2)^0.5 * d0

 

am I going wrong on the brackets or is it the units, don't know...

thanks again all

JD mather.zip