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:
 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