# Degrees and Angles

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Maths was never my forte.:confused:

Q1. If over a distance of 68 metres the land rises from 80m to 82m, what is the rise?

Q2. If over a distance of 82 metres the land rises from 80m to 90m, what is the rise?

Ben

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Nor mine Ben,

however for Q1 imagine that you are standing at the apex of a triangle. The right angle of the triangle is 68 metres away and at the same level as you. The hypotenuse slopes away from you and rises by 2 metres over the distance of 68 metres. We therefore know the length of the adjacent side (68 metres) and the opposite side (2 metres) so the tangent of the angle that you want is 2 divided by 68 or .02941 which gives an angle of 1.685 degrees

Q2 is 10 divided by 82 = 6.95 degrees.

Regards

Mike

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The correct answer is 2m & 10m.

If you mean what is the incline.......... that will be 1.685 degrees and 7.004 degrees respectively.

Mike got Q1 correct by accident. The tan-1 just happens to come out at the same number as sin-1 in this case. at school they say show your workings..... in this case not a good idea as you would have revealed your boo boo!

This is actually a sin/inverse sin problem not tan. The slope of the land is the hypotenuse not the adjacent.

Anyone care to show me otherwise...... I could be wrong, but this is something I thought I was good at in school many years ago :confused:

J

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I'll take your word for it J. I need to have a lay down after reading that.:yuk:

And thanks.....

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Now I`m sure I`ll never take up maths as a hobby.

Frank.

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The correct answer is 2m & 10m.If you mean what is the incline.......... that will be 1.685 degrees and 7.004 degrees respectively.

Mike got Q1 correct by accident. The tan-1 just happens to come out at the same number as sin-1 in this case. at school they say show your workings..... in this case not a good idea as you would have revealed your boo boo!

This is actually a sin/inverse sin problem not tan. The slope of the land is the hypotenuse not the adjacent.

Anyone care to show me otherwise...... I could be wrong, but this is something I thought I was good at in school many years ago :confused:

J

G'day J,

the slope of the land is indeed the hypotenuse. We are given the value of the adjacent side (68 metres) and the opposite side (2 metres) therefore we need to solve the angle by using Tangent = Opposite over Adjacent. We don't have the value for the hypotenuse so we can't use Sine, though being a right angle triangle we can of course solve for the hypotenuse and if we do that for Q1 we get an answer of 1.684684318 degrees. The same as we got the other way.

For Q2 if we use the same approach we get 6.952957468 degrees - the same as the other way.

This is the only part of maths that I do understand!

BTW the hypotenuse for Q2 is 82.60750571 and for Q1 is 68.02940541 and the calculation then Sine = Opposite over Hypotenuse.

Regards

Mike

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Mike

You can not be so sure of that. In fact on reflection nor can I. The question did not specify whether it was the adj or hypot being quoted.

In the absence of that I have to assume that if a human being was measuring his paddock for say an airstrip, (and given this is an aaviation forum not a surveyors or mapping one) that the person has an accurat altimeter or theodilite and measured the rise in question, and then whipped out a measuring wheel and trundled up or down the slope to measure a length.

Unless he excavated the 2m or 10m depths and measured the flat......in which case you would be correct.

Now what was the point of all this anyway???:hittinghead:

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Hi J430

I'm with you.

Unless the person measuring it was a surveyor you would have to assume that they measured the slope strip distance not usually recognising that the distance measured along a level surface would be shorter than one measured along a uniformly sloping surface.

This makes the sine function the sensible one to use in calculating the angle.

Alternatively the first example slope could be described as say 2 on say almost 68 for the first example where 68 represented the horizontal not quite the level distance and which is not true in this case.

In reality it would be 2 on slightly less than 68 as the level distance is slightly shorter than the hypotenuse even allowing for the curvature of the earth which would tend to increase it.

But we cannot measure this easily so use the 2/68 and the sine-1 function.

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Alternatively of course the person in question is not only an airstrip constructor but a modern airstrip constructor. They were concerned about a small tree, indeed in the first question a bonsai tree, infringing the approach path. They wanted to know the intrusion on the approach path and thus measured the distance, being a modern person, to the base of the tree with a laser rangefinder......

In the second question the tree was of not quite bonsai proportions and was further away, still on the flat land at the end of the airstrip....

But as it was the land rising, I will allow that there is some merit in your suggestion that it is the hypotenuse that is the valid one.

There is of course the suggestion wisely made by Ross that this is in fact not a right triangle but a spherical triangle. He suggests that we forget about that, but of course aviation is a precise business so we should really solve this triangle using Napier's rules...:big_grin:

Oh what fun.

Mike

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I knew this would get out of hand!

The engineer in me says.......2 & 7 degrees is near enough!:thumb_up:

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Is it time to re-word an 'aircraft on a conveyer belt' thread? :big_grin:

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The rise is 2m and 10m respectively. You don't need trig to work that out...

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The rise is 2m and 10m respectively. You don't need trig to work that out...

Nope you dont...... you just need to READ........see post 3

J:thumb_up:

Oops. :confused:

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The rise is 2m and 10m respectively. You don't need trig to work that out...

There's always a spoil sport somewhere!

We LIKE using our trig. Just because we weren't asked for it, doesn't mean we aren't allowed to roll it out!

Where would we all be without something to quibble about?

Kind regards

Mike

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There's always a spoil sport somewhere!We LIKE using our trig. Just because we weren't asked for it, doesn't mean we aren't allowed to roll it out!

Where would we all be without something to quibble about?

Kind regards

Mike

Or being aviators who never carry 4 figure trig tables in the cockpit we could just apply the 1-in-60 rule thus:

2 X 60/68 = 1.76 = about 2 degrees

or, 10 X 60/82 = 7.3 = about 7 or 8 degrees

Bendorn, the 1-in-60 rule provides a rule of thumb based on the reasonably accurate assumption that the sine of any angle, up to about 45Â°, is equal to 0.01666 times (or 1/60) the number of degrees. e.g sine 30Â° is 0.01666 x 30=0.5 or 30/60 = 0.5. The sine is the ratio – in any approximate right angle triangle – of the side opposite the angle, to the hypotenuse (the longest side), thus the 1-in-60 rule is very handy in the mental arithmetic of flight theory and basic navigation as the angles involved are usually less than 45Â°. For angles up to 15Â° or 20Â° the tangent (opposite side/adjacent side) is practically the same value as the sine. For angles between 50Â° and 75Â° the sine is about 1/70 times the number of degrees, and for angles between 75Â° and 90Â° the sine approaches unity.

cheers

John

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the horror....THE HORROR!!!

Thanks for all of your replys. I've now got my measurements all laid out and my inclines sorted, now I just need some extra rain to compact it down.

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All well and good for short distances, but when setting out large airstrips they have to take into consideration the curvature of the earth. I did work it out once that over a kilometer it is quite a sizeable difference from a horizontal line of sight.

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All well and good for short distances, but when setting out large airstrips they have to take into consideration the curvature of the earth. I did work it out once that over a kilometer it is quite a sizeable difference from a horizontal line of sight.

19.6 mm to be exact

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"Horizontal Line of Sight"

Hi Seal

I am not a surveyor but I have an old "Wild" diary. Wild is a firm based in Heerbrugg, Switzerland that makes or did make optical surveying equipment.

The diary has a little table in it to correct staff readings when doing horizontal levelling for the effects of curvature of the earth and refraction due to bending of light rays.

Refraction reduces errors due to the curvature of the earth. Because of refraction out here on the plains you can still see the sun for a short time after it has actually set. Your brain assumes that light travels in straight lines so you "see" the sun above the horizon.

These errors normally are not cumulative if you do levelling in a series of optical instrument set ups with equal length fore-sights and back-sights from the starting point to the end point. The curvature and refraction errors cancel out as do the instrument errers.

So you might cover a kilometre with two or three or more sets of readings with the instrument set up in the middle of each section depending on the power of the telescope in the level. Only the two reading distances at each set up need to be the same length not the lengths of each of all the reading distances.

The refraction error can vary according to the temperature of the air and it's density and moisture content etcetera so that also affects the long distance shots depending on how and where the instrument is set up.

If you only took one reading with the instrument sitting over or near the start according to their table the nominal error would be about (0.07 x Km) metre where Km is the distance of the transit expressed in Km. So that is 0.07 metre error at 1 Km or 0.27 metre at 2 km or 0.61 metre at 3 km distance plus any instrument error.

So these calculations are based on refraction of light rays for an optical instrument and an assumed radius of the earth which is assumed to be a sphere but it is not a sphere. It is a slightly squashed sphere making the correction different depending on the distance from the equator.

Apparently this is taken into account in modern GPS's that have built in tables to correct the calculated AMSL for distance from the equator and of course there is no refraction of light rays near the surface of the earth error because it does not have to connect to the starting point with an optical sighting.

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My maths must be a bit out as I get 0.291m for 1km.

I used 360*60 as the circumference of earth, giving a radius of 3434.747km, that equates to 0.017 degrees and 0.017deg for 1 km gives 0.291m. No allowance for refraction and I cannot remember the figures from my sailing days to work it out.

Now I know why I have such trouble landing, the strip is lower than it appears, at least that is my story and i am sticking to it.

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A convincing argument, I believe you!

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According to the website below

the equatorial earth radius is 6,378.137 Km

and

the polar earth radius is 6,336.75 Km

Using Pythagarus to calculate the hypotenuse then at equator for the 1 km level the curvature difference calculation

Hypotenuse = sqrt((6378.137 x 6378.137) + 1) = 6378.137078 km

Difference = Hypotenuse - earth radius there = 6378.137078 - 6378.137) Km

= 0.000078 Km

= 0.078 metre ( Curvature error at equator only based on 1 km sitting on a level line )

I wish I knew that years ago when I had to write a Fortran program to do the scoring and reporting at the Gliding comps held at Leeton with latitude and longitude of the turn points and out landing positions as the only input into distances calculated. There was no one that I knew of to help. I eventually assumed the earth was a sphere of some diameter between the diameter at the equator and the diameter at the poles which I cannot remember and then used polar geometry to calculate distances. Time input was in start and finish times in hours, minutes and seconds and had to be converted to decimal hours in the program. All results had to show the input data so it could be verified if challenged.

I just recalled I think that I would have used the formula for distance as

Distance of a leg (arc of a circle) = Rad of earth x angle in radians from start & finish point of each leg to centre of the earth

I just found a few days ago that many units of measure used would have all been redefined by agreements by international standards committees around about that time.

Not to this day has anyone wanted to argue over the results that I know of. I suppose that for most situations my errors in distance would probably not have made any difference to the final results except for minor differences in distances travelled and the speeds achieved. I think there was at least one contender for a world record speed as it was a period of intense thunderstorm activity with the safety officer threatening to resign during the competition.

That program eventually disappeared in a computer crash or a hardware or software upgrade.

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Divide by PI to get diameter 21600/Pi=6875.493542 NMiles

approximately = 6370.777312 Km

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approximately = 6370.777312 Km

phew! how many decimals would qualify for an "accurate" answer?:big_grin: