| Calibration
of clinometer test |
Robert
Leverett |
| May
23, 2007 12:11 PDT |
ENTS,
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
Then:
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.
Bob
Robert T. Leverett
Cofounder, Eastern Native Tree Society
|
| More
on clinometer calibration |
Robert
Leverett |
|
May
24, 2007 12:56 PDT |
ENTS,
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 can’t 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 don’t 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.
Bob
Robert T. Leverett
Cofounder, Eastern Native Tree Society
|
| RE:
More on clinometer calibration |
Edward
Frank |
| May
24, 2007 18:08 PDT |
Bob,
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:
http://www.nativetreesociety.org/measure/suunto_clinometer_testing.htm
Ed Frank
|
| Back
to Ed - More on clinometer calibration |
Robert
Leverett |
| May
25, 2007 08:45 PDT |
Ed,
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.
Bob
|
| More
on clinometer calibration experiments |
Robert
Leverett |
| May
25, 2007 10:54 PDT |
ENTS,
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
haven’t 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.
Bob
Robert T. Leverett
Cofounder, Eastern Native Tree Society
|
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