[photo-3d] Re: 3d equations

["bart kelsey" <attraxe@xxxxxxxxxxx>]

G'day Chris

Thanks for your prompt reply.

I substituted Z = M(sf/p - 2h) into P = (Ve)/(e-Z) and arrived at the same 
solution as yourself: P = (Vep)/(ep - Msf + 2Mhp)
Substituting h=0 into the above gave
P = (Vep)/(ep - Msf)

So, I agree with your solution, however, I find it hard to believe that such 
a mistake could have been published, particularly as it effects all the 
other equations. Is it possible that we are both wrong?

I will follow up your suggestion that an errata note may have been published 
later.

Once again, thankyou for your time and effort.

I'll let you know how I get on.

Cheers

Bart


>From: chrisp@chrisp.worldonline.co.uk
>Reply-To: photo-3d@yahoogroups.com
>To: photo-3d@yahoogroups.com
>Subject: [photo-3d] Re: 3d equations- New Scientist
>Date: Mon, 12 Feb 2001 19:26:14 -0000
>
>
>Hello Bart
>
>I now understand where your problem lies, and I now see that I had
>not spotted the problem in the article. I had just accepted that z =
>2h-(sf/p).
>
>I think you are correct that z = 2h-(sf/p) is not the case.
>
>In your calculations for z, Z and P when h>0, a negative value of
>parallax would mean that you would have to cross eyes to see the
>image - clearly not the intention of the author of the article. And
>the object in the image (P) has now shot from 6.6 metres behind the
>screen to 31cm infront, just by decreasing the inter-lens separation
>from 63 mm to 61.8 mm. Anybody experienced in stereo imaging will
>tell you this does not ring true.
>
>Thinking logically, if h is the movement of each lens inwards, then
>parallax (which begins at z = sf/p) would decrease by 2h.
>
>So surely z = (sf/p)-2h.
>
>You are correct that equation I does not work, because it is based
>upon false equations for z and Z. The correct equations are, if I am
>correct:-
>
>z = (sf/p) - 2h
>
>Z = M(sf/p - 2h)
>
>      Ve          Vep
>P = ----- = -------------       equation I
>      e-Z     ep-Msf+2hMp
>
>And likewise equations II, III, and IV  DO have to corrected, as you
>originally wrote. However, I think it would be easier to forget the
>special case of h=0 and create equations that include h and would
>work whatever the value of h.
>
>It is possible that this problem in the article has been noticed
>before. Was an errata note published in a later issue ?
>
>Do you agree with my result for equation I ? When h=0 is substituted
>into this, P=Vep/(ep-Msf) which you believe gave the correct answer.
>
>
>Chris
>
>
>
>
>
>
>--- In photo-3d@y..., "bart kelsey" <attraxe@h...> wrote:
> > Dear Chris
> >
> > Thankyou for taking the time to respond to my questions regarding
>3d
> > equations.
> >
> > You clarified the second and third questions I posed however, I am
>still a
> > bit unsure when applying equation 1. May I trouble you one last
>time to end
> > my frustration?
> >
> > I can see from the geometry where z= sf/p comes from, but not z =
>2h -
> > (sf/p).
> > I have it in my head that the total parallax should be z = 2h +
>(sf/p).
> > Could you straighten me out on this one please.
> >
> > I understand that the value of P will change depending on the value
>of h.
> > This wasn't the point of confusion. In the example of no axial
>offset I
> > substituted Mfs/p for Z in the equation P = (Ve)/(e-Z). The
>equation reduced
> > to P = (Vep)/ (ep – Mfs) which gave the correct answer.
> > However, substituting  h = 0 into equation 1 did not reduce it to P
>= (Vep)/
> > (ep – Mfs), but to P = (Vep)/ (ep + Mfs), thus giving the incorrect
>answer.
> > This is why I asked, why doesn't equation 1 work when h = 0? I'm
>still
> > unsure of this one.
> >
> > Thanks for pointing out my error in the case of axial offset.
> >
> > I look forward to a response at your convenience.
> >
> > Thanks
> >
> > Bart
>
>
>
>

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