Line of sight calculation
Guest
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You were close Diver the sq root of 1.5 is 1.22 So the formula becomes 1.22*sq root of the ht in feet.
But thanks anyway but still looking for how it was derived so come on you budding mathematicians out there Hassel may be on the right track.
Thanks in advance
But thanks anyway but still looking for how it was derived so come on you budding mathematicians out there Hassel may be on the right track.
Thanks in advance
Guest
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Isn't it a combination of Pythagoras' Theorem with an arc for the base side?
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Guest
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This is clearly due to the curvature of the earth:
Let:
Line of sight range = L
Radius of the earth = R
Altitude = h
(Units not important at this stage)
Draw a diagram of an arc of a circle with a point just above the arc to represent the position of the aircraft. Draw a line from the position of the aircraft to a tangent to the arc, then connect the aircraft and the tangent point to the centre of the arc (Centre of the earth).
We can now represent the geometry in terms of a right-angled triangle, with the right-angle being at the point of the tangent to the arc. Hence, using the notation set out at the beginning, the triangle has three sides of lengths:
L
R, and
(R + h) - This forms the hypoenuse, hence, from Pythagoras:
L^2 + R^2 = (R + h)^2
Where "^2" indicates a 'squared' value. So, expanding gives:
L^2 + R^2 = R^2 + 2Rh^2 + h^2
Where R^2 cancels out on both sides, and the h^2 term disappears since h is negligible with respect to R. (The solution becomes unnecessarily messy otherwise). Rearranging, this leaves us with:
L^2 = 2Rh
or L = Sqrt(2Rh)
Now 2R is about 7000 miles, so dividing by about 5000 <feet in a mile. to get h from feet into miles and therefore get all the units consistent <to give L in miles>, means the coefficient inside the bracket becomes about 1.5 <very roughly>, hence:
L = Sqrt(1.5h)
where h (height) is in feet and L (line of sight range) is in miles.
If you want to extend that theory to the line of sight range between two aircraft, such as the one Diver quoted then just add a similar expression for the other aircraft on the other side of the tangent point. (The two line of sight distances can simply be summed. They must form a straight line as they both hit tangents at the same point on the same curve). Just note that Diver's expression has the 1.5 coefficient outsite the square root, so it is a slightly lower number.
Right, Back to the beers then
Pie
[edited for typos]
[This message has been edited by Pielander (edited 29 May 2001).]
Let:
Line of sight range = L
Radius of the earth = R
Altitude = h
(Units not important at this stage)
Draw a diagram of an arc of a circle with a point just above the arc to represent the position of the aircraft. Draw a line from the position of the aircraft to a tangent to the arc, then connect the aircraft and the tangent point to the centre of the arc (Centre of the earth).
We can now represent the geometry in terms of a right-angled triangle, with the right-angle being at the point of the tangent to the arc. Hence, using the notation set out at the beginning, the triangle has three sides of lengths:
L
R, and
(R + h) - This forms the hypoenuse, hence, from Pythagoras:
L^2 + R^2 = (R + h)^2
Where "^2" indicates a 'squared' value. So, expanding gives:
L^2 + R^2 = R^2 + 2Rh^2 + h^2
Where R^2 cancels out on both sides, and the h^2 term disappears since h is negligible with respect to R. (The solution becomes unnecessarily messy otherwise). Rearranging, this leaves us with:
L^2 = 2Rh
or L = Sqrt(2Rh)
Now 2R is about 7000 miles, so dividing by about 5000 <feet in a mile. to get h from feet into miles and therefore get all the units consistent <to give L in miles>, means the coefficient inside the bracket becomes about 1.5 <very roughly>, hence:
L = Sqrt(1.5h)
where h (height) is in feet and L (line of sight range) is in miles.
If you want to extend that theory to the line of sight range between two aircraft, such as the one Diver quoted then just add a similar expression for the other aircraft on the other side of the tangent point. (The two line of sight distances can simply be summed. They must form a straight line as they both hit tangents at the same point on the same curve). Just note that Diver's expression has the 1.5 coefficient outsite the square root, so it is a slightly lower number.
Right, Back to the beers then
Pie
[edited for typos]
[This message has been edited by Pielander (edited 29 May 2001).]
Guest
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Let d be the distance the aircraft is away from the station,
Let a be the altitude of the aircraft,
Let r be the radius of earth (6378.15 km=3343.925 nm),
The aircraft flying at an altitude of a above the earth curved surface, the earth has a radius of r, d is the distance the aircraft is away from the station
Using Pythagorean theorem, the r^2=f^2+d^2, where f=r-a
d=sqrt(r^2-f^2)
where f^2=(r-a)^2=(r-a)(r-a)=r^2-2ra+a^2
d=sqrt(r^2-(r-a)^2)
d=sqrt(r^2-{r^2-2ra+a^2})
d=sqrt(r^2-r^2+2ra-a^2)
d=sqrt(2rb-a^2)
d=sqrt(a{2r-a})
since d and r are much greater than b, ie when a at 30000 ft = 4.99 nm so the term (2r-a), can be simplified as 2r giving an error of 0.07393 % at 30000 ft and smaller at lower altitudes
Therefore d is approximately equal to
d=sqrt(a(2r-a)) =~ sqrt(2ra)
now if a is in feet, and r, and d in nm,
d=sqrt(2*3343.925*a/6076.1154)
as there is 6076.1154 ft per nm
d=sqrt(1.1334a)
d=1.0646*sqrt(a)
If someone can post a picture on the internet site for me, I can draw a diagram of how this was derived.
[This message has been edited by Zeke (edited 31 May 2001).]
Let a be the altitude of the aircraft,
Let r be the radius of earth (6378.15 km=3343.925 nm),
The aircraft flying at an altitude of a above the earth curved surface, the earth has a radius of r, d is the distance the aircraft is away from the station
Using Pythagorean theorem, the r^2=f^2+d^2, where f=r-a
d=sqrt(r^2-f^2)
where f^2=(r-a)^2=(r-a)(r-a)=r^2-2ra+a^2
d=sqrt(r^2-(r-a)^2)
d=sqrt(r^2-{r^2-2ra+a^2})
d=sqrt(r^2-r^2+2ra-a^2)
d=sqrt(2rb-a^2)
d=sqrt(a{2r-a})
since d and r are much greater than b, ie when a at 30000 ft = 4.99 nm so the term (2r-a), can be simplified as 2r giving an error of 0.07393 % at 30000 ft and smaller at lower altitudes
Therefore d is approximately equal to
d=sqrt(a(2r-a)) =~ sqrt(2ra)
now if a is in feet, and r, and d in nm,
d=sqrt(2*3343.925*a/6076.1154)
as there is 6076.1154 ft per nm
d=sqrt(1.1334a)
d=1.0646*sqrt(a)
If someone can post a picture on the internet site for me, I can draw a diagram of how this was derived.
[This message has been edited by Zeke (edited 31 May 2001).]
Guest
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Pielander and Zeke thanks understand how the formula is derived.
Interesting point raised here is that both of you come up with a different answer Pielander with 1.5 and Zeke with 1.13.
The book that initiated the original enquiry was Thom's Air Pilot's vol 5 where he states that
"an approx max distance in NAUTICAL miles is given by the relationship
radar range = sq root of (1.5*ht AGL in feet)"
Would you agree that using a 1.5 factor gives the answer in statute miles and for nautical miles a factor of 1.13 or did I miss something
Interesting point raised here is that both of you come up with a different answer Pielander with 1.5 and Zeke with 1.13.
The book that initiated the original enquiry was Thom's Air Pilot's vol 5 where he states that
"an approx max distance in NAUTICAL miles is given by the relationship
radar range = sq root of (1.5*ht AGL in feet)"
Would you agree that using a 1.5 factor gives the answer in statute miles and for nautical miles a factor of 1.13 or did I miss something
Guest
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Engineer
Zeke is probably quite correct. I used very rough figures to fudge the answer to be about 1.5. In all seriousness, that is probably as close as you need for a 'rule of thumb' calculation. (about as accurate as "1 thumb = 10nm" sort of thing).
Close enough for Trevor Thom anyway, who is, incidentally, not my favourite author!
Pie
Zeke is probably quite correct. I used very rough figures to fudge the answer to be about 1.5. In all seriousness, that is probably as close as you need for a 'rule of thumb' calculation. (about as accurate as "1 thumb = 10nm" sort of thing).
Close enough for Trevor Thom anyway, who is, incidentally, not my favourite author!
Pie
Guest
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The old 1858 Clarke model of earth used a radius of 6378293 meters, and the international 1972 model used a radius of 6378388 meters. I approximated the radius of the earth based upon the WGS84 model of the earths ellipsoid (which is the standard for GPS and navigational aids), where the semi major axis is 6378137 meters (I approximated with 6378.15 km).
To be more accurate one should factor in the inverse flattening (1/f) 1/298.257223563 for the shape of the ellipsoid, then things get complicated when you need to know the position of the station and the aircraft to determine the shape of the geodesic curve between them. If you are going to get that accurate them you need to look at masking by terrain, and the type of terrain the radio wave are propagating over.
So depending on what model of earth you use, you will get different numbers, 40nm @1000 ft, and 50 nm @ 1500 ft are about right.
1.23*sqrt(altitude) is close enough, and is easy to learn.
To be more accurate one should factor in the inverse flattening (1/f) 1/298.257223563 for the shape of the ellipsoid, then things get complicated when you need to know the position of the station and the aircraft to determine the shape of the geodesic curve between them. If you are going to get that accurate them you need to look at masking by terrain, and the type of terrain the radio wave are propagating over.
So depending on what model of earth you use, you will get different numbers, 40nm @1000 ft, and 50 nm @ 1500 ft are about right.
1.23*sqrt(altitude) is close enough, and is easy to learn.
Guest
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P&Z
You may have misconstrued the point I was trying to make in my last post.Most probably due to my poor english. But let me have another go and then will be grateful for your comments:
Ignoring the oblate spheroid shape of the earth's I used the definition of the nautical mile to give the diameter of the earth i.e (360*60)/pi
Hence diameter = 2*radius = 6875.5nm (as an aside this equates to approx 0.2% error on the correct radius)
Now representing this figure in statute miles using conversion factor 1.15078 this results in 2r = 7912.2sm.
Armed with these two figures and using your derived equation
d^2 = 2rh having ignored the h^2 since r>>h
where
d = line of sight distance
2r = diameter of the earth
h = ht above earth's surface
Now this is the part where I may be making a mistake but can not see it. Anyway here we go.
Using this information if I want the LHS of the equation to be in nautical mile then the diameter used in the RHS must be the figure 6875.5
But if I want the LHS of the eqn in feet then this must be divided by 6076 i.e feet in a nm. Giving the result-:
d^2 = 6875.5h/6076 = 1.13h Zeke's solution
Conversely if I want the solution in statute miles must use the figure 7912.2 therefore to represent the LHS of eqn in feet divide by 5280 i.e feet in a statute mile. Hence
d^2 = 7912.2h/5280 = 1.5h Pielander's solution
also that quoted in the book I mentioned, stating that the answer was in nautical miles
In my opinion the this rule of thumb(sq root of 1.5h) pertains to a line of sight calculation in statute and not nautical miles. Hope you can concur or point out the errors of my ways Thanks a lots
(edited for typo error)
[This message has been edited by Engineer (edited 31 May 2001).]
You may have misconstrued the point I was trying to make in my last post.Most probably due to my poor english. But let me have another go and then will be grateful for your comments:
Ignoring the oblate spheroid shape of the earth's I used the definition of the nautical mile to give the diameter of the earth i.e (360*60)/pi
Hence diameter = 2*radius = 6875.5nm (as an aside this equates to approx 0.2% error on the correct radius)
Now representing this figure in statute miles using conversion factor 1.15078 this results in 2r = 7912.2sm.
Armed with these two figures and using your derived equation
d^2 = 2rh having ignored the h^2 since r>>h
where
d = line of sight distance
2r = diameter of the earth
h = ht above earth's surface
Now this is the part where I may be making a mistake but can not see it. Anyway here we go.
Using this information if I want the LHS of the equation to be in nautical mile then the diameter used in the RHS must be the figure 6875.5
But if I want the LHS of the eqn in feet then this must be divided by 6076 i.e feet in a nm. Giving the result-:
d^2 = 6875.5h/6076 = 1.13h Zeke's solution
Conversely if I want the solution in statute miles must use the figure 7912.2 therefore to represent the LHS of eqn in feet divide by 5280 i.e feet in a statute mile. Hence
d^2 = 7912.2h/5280 = 1.5h Pielander's solution
also that quoted in the book I mentioned, stating that the answer was in nautical miles
In my opinion the this rule of thumb(sq root of 1.5h) pertains to a line of sight calculation in statute and not nautical miles. Hope you can concur or point out the errors of my ways Thanks a lots
(edited for typo error)
[This message has been edited by Engineer (edited 31 May 2001).]
Guest
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Engineer,
I think your observations are valid and may very well be the case that the author of the book you were reading is not able to derive the formulae from first principles.
I checked the Australian Instrument Rating Manual (the basis for your vol 5) by Trevor Thom and it has the same formulae in it (word for word), as his company (Aviation Theory Centre) are based in Australia, they cannot use the excuse that its an American text (and using statute miles).
In Australia there are published limits for the use of VOR&DME,
<5000 (5k) 60nm
5k<10k 90nm
10k<15k 120 nm
15k<20k 150 nm
>20k 180 nm
I think your observations are valid and may very well be the case that the author of the book you were reading is not able to derive the formulae from first principles.
I checked the Australian Instrument Rating Manual (the basis for your vol 5) by Trevor Thom and it has the same formulae in it (word for word), as his company (Aviation Theory Centre) are based in Australia, they cannot use the excuse that its an American text (and using statute miles).
In Australia there are published limits for the use of VOR&DME,
<5000 (5k) 60nm
5k<10k 90nm
10k<15k 120 nm
15k<20k 150 nm
>20k 180 nm
Guest
Posts: n/a
The radio horizon is slightly more than the visual horizon because the radio waves do get bent a bit. This assumes, by the way, no anomalous propagation, such as reflections from aurora or clouds of sporadic ionisation in the E layer of the ionosphere, or tropospheric ducting.
With a temperature inversion, it's possible to get very high signal loss over a slant path of only 10,000 feet.
With a temperature inversion, it's possible to get very high signal loss over a slant path of only 10,000 feet.
Guest
Posts: n/a
The way I would see it is the formula is derived by using basic mathematical principles like Projectiles, Vectors, Pythagoras and trig.
Taking the curvature of the earth as the line of projection and taking things like aurora and propagation to be negligible, It is probably quite simple to work it out, bear in mind I have only been thinking about this for about 5 seconds, so will investigate more tonight,
Rusty
P.S Sorry If I have done a mimic of any a bove post, I haven't had time to fully read the thread.
Taking the curvature of the earth as the line of projection and taking things like aurora and propagation to be negligible, It is probably quite simple to work it out, bear in mind I have only been thinking about this for about 5 seconds, so will investigate more tonight,
Rusty
P.S Sorry If I have done a mimic of any a bove post, I haven't had time to fully read the thread.
Guest
Posts: n/a
Thanks Gentleman
I get the impression this is another one of those rules of thumb that has its roots set back in the good old days and is now used verbatim with no one questioning its origin or how derived.
Zeke
One observation you made that may hold a clue to the origin is that of the Americans using statute miles. If an American had developed it then the rule will be based on statute miles. Another option is that it deals with ground based observation so once again staute miles would be used. Who knows?
Well you learn something new everyday.
Cheers
I get the impression this is another one of those rules of thumb that has its roots set back in the good old days and is now used verbatim with no one questioning its origin or how derived.
Zeke
One observation you made that may hold a clue to the origin is that of the Americans using statute miles. If an American had developed it then the rule will be based on statute miles. Another option is that it deals with ground based observation so once again staute miles would be used. Who knows?
Well you learn something new everyday.
Cheers
Guest
Posts: n/a
<font face="Verdana, Arial, Helvetica" size="2">I get the impression this is another one of those rules of thumb that has its roots set back in the good old days and is now used verbatim with no one questioning its origin or how derived.</font>
range (units not specified) = 1.23*sqrt(alt in feet)
It is on an old handout from a CPL course in Canada and I don't know its provenance.
I have searched a variety of N American texts and can only find tables for line of sight. As expected the results from any of the formulae given, in reality are not that far apart.
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edited for for typos (again
)[This message has been edited by Code Blue (edited 31 May 2001).]
Guest
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Code Blue
Thanks for your info as you will appreciate the sq root of 1.5 is approx 1.23(to 2 DP) so the formula
range (units not specified) = 1.23*sqrt(alt in feet)
is another way of expressing the 1.5 conversion factor.
Your further info from Canadian/North American text would appear to give further credance to the theory that the range in question is in statute miles.
Maybe an aviation author out there might provide the correct answer who knows.
Another twist to the tale I suppose "is not to blindly accept what you are told" But that could be another thread to start in Wannabes.
Thanks and take care
Thanks for your info as you will appreciate the sq root of 1.5 is approx 1.23(to 2 DP) so the formula
range (units not specified) = 1.23*sqrt(alt in feet)
is another way of expressing the 1.5 conversion factor.
Your further info from Canadian/North American text would appear to give further credance to the theory that the range in question is in statute miles.
Maybe an aviation author out there might provide the correct answer who knows.
Another twist to the tale I suppose "is not to blindly accept what you are told" But that could be another thread to start in Wannabes.
Thanks and take care
Guest
Posts: n/a
Engineer,
Some other rules of thumb are at http://www.flightinfo.com/html/rules_.shtm
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It is possible to fly without motors, but not without knowledge and skill.
— Wilbur Wright
Some other rules of thumb are at http://www.flightinfo.com/html/rules_.shtm
------------------
It is possible to fly without motors, but not without knowledge and skill.
— Wilbur Wright
Guest
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Zeke
Thanks already had a look here at a few examples noticed the 1:60 rule and its derivative for the 3 degree glideslope. But no mention of the line of sight calculation. Maybe with my new found knowledge I might make a contribution.;-)
Take care
Thanks already had a look here at a few examples noticed the 1:60 rule and its derivative for the 3 degree glideslope. But no mention of the line of sight calculation. Maybe with my new found knowledge I might make a contribution.;-)
Take care
Guest
Posts: n/a
Starting with an equatorial dia of 12,756 km, I got
d=1.13 * sqrt(h)
using the same pythagorean approach as you all did.
Then I went to Norie's Nautical Tables (1969) page 149 for a table "Distance of Sea Horizon for given heights".
That table gives 10' 3.63nm
100 11.5
1000 36.3
5000 81.3
10000 115
Back-calculating the elusive coefficient everyone is arguing about, all give 1.15, and notes confirm that the table is derived from d = 1.15 sqrt (h). No reference is quoted. He does mention provision being made for atmospheric refraction.
Don't know about heights over 10,000ft, but it seems to me that for practical purposes, aviators could use anything from 1.1 to 1.5. Atmospheric haze is going to restrict you from practical checks.
The same can not be said for old time sailors with a lighthouse in view. Light heights were always carefully reported on charts so that this rangefinding method would give you the best circle of position possible.
ET
d=1.13 * sqrt(h)
using the same pythagorean approach as you all did.
Then I went to Norie's Nautical Tables (1969) page 149 for a table "Distance of Sea Horizon for given heights".
That table gives 10' 3.63nm
100 11.5
1000 36.3
5000 81.3
10000 115
Back-calculating the elusive coefficient everyone is arguing about, all give 1.15, and notes confirm that the table is derived from d = 1.15 sqrt (h). No reference is quoted. He does mention provision being made for atmospheric refraction.
Don't know about heights over 10,000ft, but it seems to me that for practical purposes, aviators could use anything from 1.1 to 1.5. Atmospheric haze is going to restrict you from practical checks.
The same can not be said for old time sailors with a lighthouse in view. Light heights were always carefully reported on charts so that this rangefinding method would give you the best circle of position possible.
ET