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Hi Jamie & All: Feb 10, 2007
If this visible object has sufficient known or measurable
dimensions, normal to your line of sight, then distance will be
dimension divided by the Tan of the angle that just includes the
dimension. In practice, measurement of small angles is an exercise in
error, so a scale model of the configuration would be more precise.
A key part of making such a model would be a good sight. To get a
grasp of proportions, I made one out of two milk carton caps (plastic, 3
cm diam) as follows: 5/64" hole drilled in center of one for a peep
sight, secured with masking tape to edge of a narrow board with hole 14
cm from end (with hand against cheek & board held between thumb and
finger it can be held steady), 3/16" hole punched in center of the
second cap with common pins stuck through cap for cross hairs and one
pin ~3.5 cm long so it sticks beyond cap for mounting, cross hairs
mounted on board edge 100 cm from peep sight. This would make a good
Squirrel rifle if the board could be loaded.
To describe in concrete terms how one might use this sight to
construct a model--- Assume dimension to be horizontal, say two bright
orange tarpaulins hung 300 metres apart (W1) on the beach and on a line
normal to the line of sight. Secure a board, aimed at and just below the
tarps and fasten the peep sight to one end. Mark the position just below
the peep hole. Draw a line on board, at a right angle to line of sight.
While looking through the peephole, say at the left tarp, have an
assistant move the cross hairs along the line just drawn until the tarp
is lined up on the sight and then push the sight pin into the board to
mark the position. Repeat for the riight tarp. Measure the distance
between the cross hair marks (W2), measure the distance (D2) between the
peep hole mark and cross line (midway between the cross hair marks).
The distance to the beach (D!) is then equal to W1 x D2/W2, all in
the same units.
Yours truly, Dave Webster, Kentville
Jamie Simpson wrote:
> Peter
>
> Are you aware of a method of determining distance of a visible object
> other than actually measuring or using a transit?
>
> If I could do that I think I could figure it out mathematically. But
> I don't think it can be done.
>
>
> ----- Original Message ----- From: "Peter Payzant" <pce@accesswave.ca>
> To: <naturens@chebucto.ns.ca>
> Sent: Thursday, February 08, 2007 12:38 PM
> Subject: Re: [NatureNS] Determining Elevation
>
>
>> Hi, Jamie-
>>
>> Elevation measurements from GPS are significantly less accurate than
>> lat-long measurements. This is due in part to the geometry of the
>> situation, and in part due to the fact that the GPS system calculates
>> your distance from the centre of a sphere, and then applies a model
>> of the earth's surface to correct for the earth's radius at your
>> location. There are other sources of error as well.
>>
>> There's lots about this on the net of course. Until you understand
>> the nature of the errors it's probably not a good idea to put too
>> much faith in what the GPS receiver tells you for elevation.
>>
>> If you wanted to get a good result, you could rent a surveyor's level
>> and work your way up to your house from the nearest known accurate
>> elevation.
>>
>> Peter Payzant
>>
>>
>> --
>> No virus found in this incoming message.
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>> 2/7/2007 3:33 PM
>>
>>
>
>
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