Conference unifix::sailing

What equipment am I allowed to used?

Radar would tell you the distance, so would an optical rangefinder (good
ones available surplus from any shipyard breaking up naval vessels -- 
but they might be a little too big for the average yacht). A laser 
rangefinder (such as used on army tanks) would also do the job nicely. 

Failing all this, get out your handbearing compass. Take a bearing. Sail 
on for a while. Take a bearing. Solve the geometry problem (two angles 
and a side -- using a handheld calculator would make it easier).
Probably won't be very accurate. 

>    Does anyone know how to determine the distance to an object of known
>    height _before_ the base of the object is over the horizon?

This is a simple geometry problem. Measure the angle a subtended by the object.

distance = tan(a) * height. 

This is the exact (okay, almost exact) solution for the distance to the base.

If the angle is small (less than 5 degrees) a useful approximation is that

tan(a) ~= a  (a is in radians, 3.14159 radians = 180 degrees)

This results in 

distance ~= a * height

To make this work, you must have a gizmo for measuring the angle
between the top and base of the object. One such gizmo is a sextant. Or, you
can make your own gizmo by putting a scale, marked in thousandths of radians,
at the end of a stick. (The radians are measured relative to the other end
of the stick.) This is basically a protractor that spans only a few degrees,
with the "center" of the protractor at one end, and the scale at the other
end. Sight down the stick from the "center" end and read off the radians.

Just multiply by the height to get the distance!

Making the scale is easier than it sounds: Use a stick that is 1 m long. Mark
the scale at 1mm intervals. Those are 1/1000 radians. 

--RS

    
    what I meant was, if you could see only part of an object above
    the horizon, but knew it's true height, how would you calculate
    the distance to it only from it's apparent height.
    
    e.g. you're sailing towards Block Island from Cuttyhunk and can
         see only part of the cliffs.  you hold a ruler up at arm's
         length, which you've previously measured to be 28" from your
         eye, and observe that B.I. measures .3" when you hold the
         ruler by its bottom.  From your chart, you see that the
         true height of B.I. is 218' (let's say), and your eye is at 
         a height of 6'.
    
         because you can't see the base of B.I., you can't use
         simple triangulation.  Because you're not bobbing B.I. over
         the horizon, you can't use the equation in .0, at least not
    	 without some modification.
    
    so, re .1, what equipment?  A ruler and a chart.  You can only make
    one observation.
      

    
    re: .3
    
        (notes .3/.4 entered at the same time)
    
        that'll work when you can see the base, but when you can't see
        the base, but can see part of the top, you don't know what the
        true height of the visible part of the mark is.
    

re .4:

If all you're allowing is a chart and one observation, you have insufficent 
data to solve the problem.

    >> If all you're allowing is a chart and one observation, you 
    >> have insufficent data to solve the problem.
    
    Can you support this?  It seems to me that if the curvature of
    the earth is properly taken into account that sufficient
    information is available.  The apparent height of the the mark
    increases monotonically as you approach it, and there is a 1-1
    correspondence between distance and apparent height.  For an
    object with a known true height, I believe this means that
    sufficient data is available.
    
       /Jim
    

    
    Does anyone know how to determine the distance to an object of known
    height _before_ the base of the object is over the horizon?  In
    general the horizon is at a distance 1.14 * (sqrt height), so if
    it is within your own horizon, or if you're just 'bobbing' it over 
    your horizon you can use this formula, but at intermediate points I've
    never heard of a way to calculate the distance.  
    
Now that I understand your question...

If you have a sextant and can accurately measure the angle subtended by
the visible part of the object, you can approximate the distance as follows:

at_horizon_angle = object_height / your_horizon_distance

percent_of_object_that_is_showing = 
	sqrt(visible_subtended_angle / at_horizon_angle)

object_distance_from_your_horizon =
	1.14 * sqrt(object_height * (1 - percent_of_object_that_is_showing))

total_distance = your_horizon_distance + object_distance_from_your_horizon


step 2 is an approximation, and it don't know how accurate it is. (I believe
it is quite accurate if your_height >> object_height :-) I believe that the
given equation is an upper bound, and that 

percent_of_object_that_is_showing = visible_subtended_angle / at_horizon_angle

is a useful lower bound. You can use the two values to bracket the actual
answer. I don't have the trigonometric stamina to derive these in detail, or
to derive a more accurate approximation for step 2. The approximation in
step 2 is derived from the following observation:

An observer far out in space observes an object at the earth's horizon. The
earth turns away at a constant rate. The apparent height of the top of the
object recedes as the square of the rotation for small rotations.

--RS

	Well you can make a SWAG.

	You know the range at which you can just see the top of the island.
	Call it R1.
	You know the range at which you can just see the base of the island.
	Call it R2.
	You know the angle to the top of the island when you can just see
	the base.
	Call it A2.
	Measure the angle to the top.  Call it Ax.

	Now the SWAG at the range (call it Rx) is:

	Rx= R2 + (R1-R2)*(Ax/A2)

	This is not exact but close and much better than nothing.

	Al

re .7:

You gave the reason yourself in .5 -- unless you know the height of what 
is appearing over the horizon, you have insufficient data to calculate 
its distance. All you can measure is the angle from the horizon to the 
top of the object. Not enough to determine the distance (sophmore high 
school geometry). Of course, you can GUESS the height, and then you can 
guesstimate the distance. Not very accurate, but better than nothing.

re .8:

Right, just measure the angle subtended by the object. If H is height 
of the object and A the angle, the distance D to the object is 

                   D = H/TAN(A)

It is difficult to get an accurate answer this way, however. Assume that 
H is 100 feet and you are five nm away. A is about 0.19 deg (11 
minutes). Small sextant angles are hard to measure. If you make a 1 
minute error in measuring (very easy), you will calculate that you are 
5.5 miles away. 

Hmmm, what happened to the previous reply .11? It was on the right 
track.

For a given height, the horizon is 1.14*sqrt(height) miles away. If an 
object is visible, the following equation may be used:

       H = 6076*1.14*(sqrt(EH)+sqrt(H))*tan(A)

where H is the object height in feet
      EH is the eye height of the observer in feet
      A is the angular height of the object above the horizon (in 
        degrees and measured by a sextant)

So, program your favorite programmable calculator to solve the equation 
by guessing the answer, and use your sextant to measure A. 

Should work ...... but reality intrudes. Assume EH = 9 feet and H = 100
feet. Then A = 3.82 minutes of arc. With an inexpensive sextant and even 
a small sea running, any measurement of A to within 5 minutes of arc is 
virtually impossible. (A study by some British institute of navigation 
showed that the probable position error using sextant sights taken by 
professional navigators from the bridges of large ships was on the order 
of 5 miles, ie, 5 minutes of arc). 

So, the method seems to work (my apologies for assuming the problem is 
insoluble) in theory, but probably isn't useful in reality. Still, the 
author of the base note wanted a method using a chart and ruler. This 
method requires a sextant and much tedious calculation (or a 
programmable calculator).

    
    >> You gave the reason yourself in .5 -- unless you know the
    >> height of what is appearing over the horizon, you have insufficient
    >> data to calculate its distance. All you can measure is the angle
    >> from the horizon to the top of the object. Not enough to
    >> determine the distance (sophmore high school geometry).
    
    That's not what I said.  What I said was that if you can't see the 
    base, then you can't use the trigonometric method described in .3.
    
    In fact, there *is* enough information.  Paul Barford gets the credit
    for puzzling out the following relationships:
    
    	Let the true height of the visible part of the object be V,
        and the invisible part be I.  If the total height is H, then
        H = V + I.
    
        Let d be your horizon distance, and A be the angle between
        the horizon and the top of the object.   Then:
    
    		V = 6038 tan A (d + 1.14 * sqrt (H - V))
    
        This is somewhat messy since V appears on both sides of the
        equation, but there is only one unknown, V.
    
        We also know the apparent height h of the object at our horizon
        distance d at angle A,
    
    		h = d tan A
    
        We now have similar triangles:  h/d = V/D, or D = V*d/h, where
        D is the linear distance to the apparent base, or:
    
    		D = V*d/h
    		  = V/tan A
    		  = (tan A * (d + 1.14 * sqrt (H - V))/tan A
    		  = d + 1.14 * sqrt (H - V)
       
        With known H and d, and solving for V with the recursive
        definition given above, you can calculate D.
    
	For the Block Island problem:
    		d = 1.14 * sqrt(6ft) = 2.79nm
    		H = 218
    		V = 217.74
    		D = 3.37nm
	     
        The equation for V did not converge all that easily for V so
        close to H.  For other angles:
                                       
    		tan A = .36/28 -> D =  2.79nm (base just visible)
    		tan A = .3/28  -> D =  3.37nm
    		tan A = .25/28 -> D =  4.02nm
    		tan A = .2/28  -> D =  6.25nm
    		tan A = 0/28   -> D = 19.62nm (top just visible)
    /Jim
    

    
    re .10 and .11
    
       I deleted and reinserted .10 because I initially left out the
       6030 ft/nm conversion factor in the equation for V.  I also
       wanted to solve the puzzler listed in .4
    
       Well, a ruler is all that's required, though that's not how
       I'd do it if I owned a pair of TASCO binoculars.
    
    	;)
    

    
    re: .12
    
        I apparently made a miscalculation for tan A = .2/28.  Using
        a programmable calculator I get the following values:
    
    		tan A = .30/28 -> V = 217.743 -> D = 3.37
    		tan A = .25/28 -> V = 216.827 -> D = 4.03
    		tan A = .20/28 -> V = 214.331 -> D = 4.98
    
        /Jim
    

    
    You know, Alan, I wanted to leave things be, but after rereading
    your reply I see you still need convincing:
    
    >> Should work ...... but reality intrudes. Assume EH = 9 feet and H = 100
    >> feet. Then A = 3.82 minutes of arc. With an inexpensive sextant and even 
    >> a small sea running, any measurement of A to within 5 minutes of arc is 
    >> virtually impossible. (A study by some British institute of navigation 
    >> showed that the probable position error using sextant sights taken by 
    >> professional navigators from the bridges of large ships was on the order 
    >> of 5 miles, ie, 5 minutes of arc). 

    reality #1:
    
    	in the problem definition, the tangent of the angle is measured
    	as .3/28, or A = 36 minutes of arc.  This is much larger than
    	the example you created.
    
    reality #2:
    
    	in real life, BI SE light is 201', so at this angle the real
        BI would be 3.07nm away, and would nearly be at our horizon
        distance.
    
    reality #3:
    
    	the taller the observer height and the shorter the object
        height, the less useful it is, since the viewing angle
    	diminishes.  This places a limit on minimum object height.
    
    reality #4:
    
        you do not need a programmable calculator, in practice, if
    	you prepare a table of distances which is indexed by viewing
    	angle and true object height.  Since the number of useful marks
    	on which this technique can be used in any given area is
    	normally very small, this can be a small table.
    
    	on the other hand, I'll grant you that having this table written
    	on the margin of the chart was not part of the problem description,
    	and it would be virtually impossible otherwise.  However, the
    	spirit of the question is to limit your *measurement* devices;
    	I also did not stipulate that you may have tables of tangents,
    	sines, etc. on board, or be able to calculate a square root.
    
    perception #1:
    
    	It seems to me that several of your replies, Alan, were unnecessary
    	and somewhat offensive.  On this point, I hope I'm wrong.
    
    /Jim
    

re .15:

Last things first: I'm sorry you found my replies offensive. They 
weren't intended to be.

However, since you would like to continue the discussion, let's 
continue. If you propose a general purpose navigational technique, then 
I think that you have the responsibility to indicate when it is useful 
(gives reasonably accurate results) and when it isn't useful (gives 
inaccurate results). In your base note you didn't indicate that you were 
looking for a techique that worked only in limited circumstances. 

The numbers I used are about what you'd have if you could see the upper 
half of the cliffs of Block Island. This would certainly be a situation 
where you'd like to know how far you are from the cliffs. It is also a 
situation where is it highly unlikely that you could get reasonably 
accurate results using the technique being discussed.

>>>    reality #1:
>>>    
>>>    	in the problem definition, the tangent of the angle is measured
>>>    	as .3/28, or A = 36 minutes of arc.  This is much larger than
>>>    	the example you created.

This is still a very small angle. Try standing on the deck and try 
reading a ruler held at arm's length to the nearest 0.1 inch. Your 
chances of making an accurate reading aren't very good. After all, a 0.1 
inch error is 33% of 0.3. If you measure 0.2 inch, you'll calculate 
you're 5.02 miles away. If you measure 0.4 inch, you'll calculate you're 
2.51 miles away. This isn't what I would call a precise navigational 
technique. The error is enough that depending on it for safe navigation 
would be less than wise. (The technique of ruler at arm's length could 
just as well be used to measure the altitude of the sun for celestial 
navigation. Sextants are used because their accuracy is vastly better.)
If you're going to propound a navigational technique, you owe it to the 
unwary to investigate and report on the accuracy of what you're 
proposing.

>>>    reality #3:
>>>    
>>>    	the taller the observer height and the shorter the object
>>>     height, the less useful it is, since the viewing angle
>>>    	diminishes.  This places a limit on minimum object height.
    
No, the limit is on the distance at which the technique is usable. The 
technique depends on both observer height and object height. 

>>>    reality #4:
>>>    
>>>     you do not need a programmable calculator, in practice, if
>>>    	you prepare a table of distances which is indexed by viewing
>>>    	angle and true object height.  Since the number of useful marks
>>>    	on which this technique can be used in any given area is
>>>    	normally very small, this can be a small table.
    
Oh, how are you going to prepare the table? Doing even a small table 
would be incredible tedious without a calculator or a computer.

My perception #1:
    
It seems to me that you knew the answer ahead of time and were trying 
see how dumb the rest of us are. On this point, I hope I'm wrong.
You also kept changing the rules of the game. Not very fair.
    
The fourth solution I offered in .1 is a technique that offers better
accuracy than what you've proposed, does not require knowledge of object
heights (which often aren't on the chart), doesn't require that you know 
for sure what you're seeing, requires no calculation (just plot on your
chart), and works anywhere you might cruise. You seem to have dismissed
it because it wasn't the answer you were looking for. 

Alan

    
    No, I didn't know the answer before I asked.  In fact, I credited
    Paul Barford with the solution, and have promised to take him
    sailing with me this summer in payment for his assistance (poor
    wretch). 
    
    I'm not proposing that everyone carry rulers in order to measure
    distances to objects.  Clearly the more accurately you can measure
    small angles, the longer the range over which the technique could
    apply.  There is the "puzzler" aspect of the question which sought
    to establish the geometrical relationships, and the "proposal"
    aspect, which I have not yet discussed.  For the purposes of the
    original puzzler, all measurements were perfectly accurate, there
    was a dead calm, and the visibility was outstanding.  Originally
    there was debate about whether there was sufficient information,
    and now that that's been established, we can move on to how useful
    or non-useful it might be.
    
    The best use for the technique may be to establish bounds on the
    range based on an analysis of the measurement errors involved, and
    combine this with a measurement of the bearing and depth.  The
    range may therefore be useful to disambiguate depth readings.
    
    e.g., if at a bearing of 090, range [3.3 : 3.5]nm, there is only one
          charted depth corresponding to the current depth D, then you
    	  could obtain a fix even from an object which was partially 
    	  obscured by the horizon.
    
    if the depths rose smoothly and monotonically this would not be
    required, since the depth and bearing would be sufficient.
    
    If mountains with visible peaks were nearby, this would allow you
    to use them for navigational purposes, even if their bases were
    well beyond the horizon.  This is the only technique I know of
    which would allow a single distant peak to be used to estimate
    a range.  
    
    btw, yes, the table I mentioned in .-2 would need to be computed by a
    program, but once computed could be placed on your chart or Eldridge
    or what have you.  If there are only 2 or 3 objects on which this
    technique would apply where you are cruising, then the table would
    pre-calculate the distances to these objects over the range of 
    viewing angles which your instruments are capable of measuring,
    along with the measurement error at each angle.
    

    /Jim
    

RE:.16 by MSCSSE::BERENS "Alan Berens"

> This isn't what I would call a precise navigational technique. The error 
> is enough that depending on it for safe navigation would be less than wise. 

Precise?  No method is completely accurate!  Safe navigation depends on 
never trusting any one piece of data completely.  Remember all numbers have 
error.  Having a different method with different sources of error can wake 
you up sometime when you need it.


{Calculating the table}

See "American Practical Navigator" table 9.  Reinventing the wheel is work!


Phil