Vertical angle

Your Kidding, Right?

Hint, there are 360°00'00" in a full circle.

Paul in PA

Your Kidding, Right?

> Hint, there are 360°00'00" in a full circle.
Cranky pants.

> can anyone share the tips and trick or the easy way to find the means value of vertical angle?
>
> case : we have two values of vrtcal angle which are
> face left : 89 36 50
> face right : 270 23 27
>
Subtract the face right value from 360°00'00".

ie/ 360°00'00" - 270°23'27" = 89°36'33"

then average the face left value and the reciprocal face right value to get 89°36'41.5"

Your Kidding, Right?

Why would you belittle a person asking a legit question?

I've always thought of the vertical angle as being the angle off the horizon.

It's how we did calculations years ago.

So the horizon is at 90 direct and flipped at 270.

Between 0 and 90 means you are looking up hill, so you have an angle of 0d23'10" in direct and 0d23'27" in reverse.

The mean is 0d23'18.5", using that angle you can apply sin, cos and tan to find your horizontal distance, and vertical distance, although to the vertical you need to apply a further correction called curvature and refraction.

Your Kidding, Right?

Your Kidding, Right?

YOU'RE kidding, right?

Your Kidding, Right?

Because he is Paul in PA. Nobody is exempt from his continual attempts to show how smart he is, not even the NGS Chief Geodesist.

😉

Your Kidding, Right?

> Because he is Paul in PA. Nobody is exempt from his continual attempts to show how smart he is...
>
> 😉

I thought it was just me that thought that..Huh, who knew? :snarky:

Actually we all typically measure zenith angles.

"A" being a vertical angle, measured from the horizon.

> > can anyone share the tips and trick or the easy way to find the means value of vertical angle?
> >
> > case : we have two values of vrtcal angle which are
> > face left : 89 36 50
> > face right : 270 23 27
> >
> Subtract the face right value from 360°00'00".
>
> ie/ 360°00'00" - 270°23'27" = 89°36'33"
>
> then average the face left value and the reciprocal face right value to get 89°36'41.5"

An easier way to understand the concept is that you are measuring two angles of a circle. Add the angles up and theoretically you will have 360° (say 90° + 270° = 360°). Any amount you are over or under 360° is your error.

Assuming you have a calculator that will add and subtract in hour-minute-second format...

89-36-50 + 270-23-27 = 360-00-17 (17 seconds of error)
Since your over 360°, subtract half that error from your initial direct reading.
If you were under, you would add half the error.

Answer is 89-36-41.5

BONUS TIP: Take the COS of that mean zenith angle (converted to decimal degrees) --> x slope distance --> + instrument HT --> - height of target = vertical difference between stations.

Pretty handy for running trig levels.

Yes, but the tables only went to 45 degrees so when the curta cranked up you needed the horizon angle. Also no computer was needed all the calculations are done in your head in a couple of seconds before getting to the table....;-)

There usually is always a reason for the way things used to be done.

[sarcasm]you just have to talk to an old timer to find out![/sarcasm]

Face left and face right are not terms used with vertical angles...

Why not, and what terms do you use?

Direct and reverse.

I've always used direct and reverse.

> I've always used direct and reverse.

I thought direct/reverse and face right/face left were interchangeable. It's more of a regional thing.....at least that's what I thought.

Now this is the kind of stuff that has fascinated me for years about the area where math and surveying meet.

Norman and Digger both get the same 17 second difference, but they do it in different ways and, of course, get the same average value. But, for the height difference, Norman would use sine where Digger uses cosine.

MightyMoe adjusted his workflow for tables that went only to 45 degrees, but that's as far as they needed to go. That's because the sine of 50 degrees is the same as the cosine of 40 degrees, as is true for all complementary angles. I wonder if his tables had sine, cosine, tangent, and cotangent as column labels across the top and cosine, sine, cotangent, and tangent across the bottom. If so, then degrees from 0 to 45 would be in the left-most column and 45 to 90 in the right-most column. Read down for the smaller angles and up for the larger ones.

In my last semester as a trig teacher, I continued to introduce my students to a similar table, not for calculation purposes, but for understanding. They could "see" the functions of angles and their complements in the tables as a supplement to seeing them in right triangle figures. That helped to remove some of the "black box" nature of their calculators and reinforce the right triangle angle relationships, as well as the constancy of trig functions of a particular angle.

Old guys who calculated mentally and by hand and then learned calculators and computers had a distinct advantage.