Re: Re: 3rd ave sensor design

From: Ken Poulton (poulton@zonker.hpl.hp.com-DeleteThis)
Date: Thu Sep 29 1994 - 14:07:38 PDT


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Date: Thu, 29 Sep 1994 14:07:38 -0700
From: Ken Poulton <poulton@zonker.hpl.hp.com-DeleteThis>
Message-Id: <9409292107.AA18291@zonker.hpl.hp.com-DeleteThis>
To: wind_talk@opus.hpl.hp.com-DeleteThis
Subject: Re: Re: 3rd ave sensor design


> Uh, folks, what's with the laser? The spinning white-and-black thingy
> seems to be to be a *much* easier deal with much lower "environmental
> impact". Lasers are sexy but focusing a telescope onto a 6"x4" target
> from a 1/2 mile (how far is the channel marker from shore at 3rd?) seems
> to me to be just as hard as hitting the same target with a laser. Even
> if you could hit it with the laser what about detecting it? How much
> power are you going to have to emit to get it out and back?

The point is that even a defocused low-power laser is much brighter from
a mile away (the distance at 3rd) than a white sunlit card. To get a
good signal from a sunlit card, you need to view *only* on the part that
gets obscured by the spinning cups. Otherwise, the background
contributes such a high ambient light that you cannot get your signal
out. (Hitting a six inch target at one mile means aiming to better
than .01 degree. This sounds hard/expensive.)

With a laser, the apparent brightness at the wavelength of interest can
be much brighter than the background, so the focusing/aiming requirement
may be relaxed. system, you need a single photodetector device (no
CCDs).

[ I just went and consulted a friend with some optics experience. ]

My friend says we need to recieve around 1 uW. If we assume a 5 mW
laser (pretty common), and assume a cheapy 3" telescope to receive with,
we need 1 uW in 7 in^2. This says we can let our 5 mW spread out
over 35000 in^2, or a 17 foot diameter. At 2 miles roundtrip, that
allows a .1 degree spread (and aiming). He thinks that is tight, but
not very hard to do.

By comparison, a sunlit card 6" on a side might reflect 11 W, but
diffusely. With 1/r^2 radiation, I think we get 98 nW/ft^2 at one mile.
Then to collect 1 uW, we need a 3.6 foot diameter telescope (and .005
degree aiming). Maybe there's one on Mt Hamilton we can use for this
during the day. :-)

Sorry for the english units, and there are plenty of losses I have
omitted, but I think these are good order-of-magnitude numbers.
I welcome further refinements by anyone with more experience.

(Kirk, I kept this idea to myself for six months before deciding *I* was
never going to have time to build this thing. If Windsight is willing
to put up the $$ to build this it and the time to get it sited, and
the $$ to run the thing, they deserve to be paid for it. From my view,
it will benefit us all if we can get better wind data, even for money.
We may as well see if we can sketch out the design for them. Otherwise,
it will probably never happen.)

Ken Poulton
poulton@opus.hpl.hp.com-DeleteThis

"Contrariwise", continued Tweedledee, "If it was so, it might be;
and if it were so, it would be; but as it isn't, it ain't. That's logic."
                                        -- Lewis Carroll



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