Calibration of clinometer test   Robert Leverett
  May 23, 2007 12:11 PDT 


       Frequent testing of the calibration of one's instruments is the
hallmark of a good Ent. The vertical side of a building provides a means
for checking the calibration of a clinometer. The following test assumes
that your laser is in good working order and accurate to within a few
inches. We'll deal with where that isn't the case and increasingly
complex situations, but for the present, let's just consider the angle.

      From a distant point, ideally on a hill where part of the vertical
side of a building is higher than your head and part is lower,
measurements and calculations are taken and performed based on the
following definitions.

Let:    Am = angle measured with clinometer to a point on building
              above eye level
        Bm = angle measured with clinometer to a point on building
               below eye level
        La = distance measured to point on building above eye level
                using laser
        Lb = distance measured to point on building below eye level
                using laser
         A = computed angle above eye level
         B = computed angle below eye level
         D = measured level distance to side of building using laser
         ea = error in angle at point on building above eye level
         eb = error in angle at point on building below eye level
         e = difference of errors
          A = arccos(D/La)
          B = arccos(D/Lb)
          ea = Am-A
          eb = Bm-B
          e   = ea - eb

       If e is 0 or nearly so, but ea <>0 and eb <>0 then the assocaited
height errors that would be made will be very nearly offsetting when
using the sine method, but not necessarily so with the tangent method as
we've seen in prior e-mails.

       This test does not show what is happening over the range of
angles that you would be using the clinometer. But this simple test can
provide some initial input on whether your clinometer is shooting high
or low. A more sophisticated experiment might test at known height
intervals on the building such as from window sill to window sill. I
recall Lee Frelich describing calibration tests for clinometers using
such a strategy.

        In the first of the above formulas, note that A = arccos(D/La)
is equivalent to computing D = La[cos(A) ] from measured values of La
and A, except in this case we know La and D from laser measurements. The
acos function allows us to then determine the angle that gives the
lengths D and La. Arcos is also called the inverse cosine function.

        In a future e-mail, I'll outline a more sophisticated experiment
that doesn't take too much time or calculations to perform. I fully
understand that the more involved these processes are the less inclined
we all are to use them.


Robert T. Leverett
Cofounder, Eastern Native Tree Society
More on clinometer calibration    Robert Leverett
   May 24, 2007 12:56 PDT 


     A "slight" flaw in the experimental design of the clinometer test
is that we have not pinpointed where exactly is level on the building,
since that canít be readily determined with an out of calibration
clinometer. However, the horizontal distance error made by misjudging
the eye level point is usually insignificant and much less than the
range of error attributable to the laser, at least for angles of a
degree or less. For example, suppose our clinometer reads 1 degree high.
Presumably, we donít know this, but we want to perform a calibration
test on our clinometer. We sight in on what we think is eye level on the
building and target that point with the laser. Hitting the presumed eye
level point (actually 1 degree high) with a perfectly accurate laser
when the distance to the building exactly at eye level is 150 feet would
produce a laser-measured distance of 150.02 feet, which compared to 150
is plenty close. So, D in the formulas A = arccos(D/La) and B =
arccos(D/Lb) will be a sufficiently accurate approximation of the exact
distance unless our clinometer is off by multiple degrees.

    As an example of this clinometer test, suppose we shoot at a point
on the side of a vertical-sided building that is actually at 45 degrees
above eye level, but our clinometer reads 46 degrees. The laser distance
to the point, which we are assuming is correct, will read 212.1 feet. If
we choose to compute the horizontal, eye level distance to the side of
the building based on the chosen point, we would calculate 212.1*cos(46)
= 147.34 feet. The difference between this calculated distance and 150
feet is 2.66 feet. This definitely tells us that our angle is in error.
But, how much is the error? This is where the formula

               A = arccos(D/La)

      comes in handy. A= arcos(150/212.1) = 44.99 - or practically
speaking 45 degrees. But our clinometer read 46 degrees. Our test would
have revealed the 1-degree clinometer error.


Robert T. Leverett
Cofounder, Eastern Native Tree Society
RE: More on clinometer calibration   Edward Frank
  May 24, 2007 18:08 PDT 


Bob sure you can tell where exactly level is on the building. Shoot
from the building to some point on a tree or post to that the clinometer
reads as level. Mark your position on the building from where the shot
was taken. Go to the secondary post or tree and shoot back at the
building from the post or tree point the first reading indicated was level. 
Have someone make a mark at the point the clinometer reads as
level on the building. If the original point and this second point do not 
line up then the clinometer is out of alignment. The exact level point is 
half-way between the the original shot point and the second point on the
shoot-back. If the shoot-back point is below the original point, then
the clinometer is reading high,   The angle it is off will be
over-reading by arc tan [1/2 (error)/distance].   If it is pointing
higher than the starting point, then it is reading low. The
calculations of error are the same.  See also: 

Ed Frank

Back to Ed - More on clinometer calibration   Robert Leverett
  May 25, 2007 08:45 PDT 


I now recall you describing this clinometer calibration test in a past
e-mail on spelunking. I read it, but didn't take it to heart. This time
I drew a diagram for myself to conform with your explanation in order to
get a better feel for how it works and I quickly convinced myself of the
value of the test. The assumption (correctly made) is that the distance
measured from the midpoint of the error segment back to the spot on the
tree/post marked as being at 0 degrees is a level line. This allows the
formula AE = arctan [1/2 (error)/distance] to be based on a right
triangle. Quite clever.

More on clinometer calibration experiments   Robert Leverett
  May 25, 2007 10:54 PDT 


    In the previously presented formula A = arccos(D/L), we assume the
distances D and L are correct. Remeber that L is the distance to the
target and D is the level distance to the eye level point vertical
beneath the target, presumably on the trunk or side of a building. If
so, then A is the angle needed to satisfy the distances D and L in the
constructed right triangle and A = arccos(D/L) is the formula needed to
return the angle between D and L. If the angle computed from this
formula differs from that read from the angle scale of a clinometer,
then we conclude that our clinometer is not registering accurately. Key
to the configuration of measurement points is knowing that the target
point is directly above the base point and the base point is at eye
level, i.e. that we truly have constructed a right triangle. If we
havenít then, we can't use this method to calculate angle error. For
example, if the target point is horizontally closer or farther than the
base point, then we have a crown-point offset situation. We will not
have an apples-to-apples counterpart to the laser-measured distance D.
It is always critically important that we fully understand the
configuration of our points.


Robert T. Leverett
Cofounder, Eastern Native Tree Society