How far (distance) can we see? What formula to use.

it may not a difficult question. see left pic of earth.

as the earth is a elliptic, north to east is its short axis and east ot west is long axis. we can see longer distance from point P than do from E point.

from ellips equition, we can get its cut line at efvery point by derivative. at the point you are standing, get its normal line. you get a triangle. now you know its tangent and normal and their clamp angle. so you can easily get their subtense line long. and corresponse arc long.

Thank you. Glad it is not a difficult question. Since it is easy for you, would you please include the formula so I can just substitute numbers to get the answer? Example:

I can see a ship at sea 50 miles away. How high above the ground do I have to be to be able to see that ship? Show me with a formula how you figure how high I am.

2 feet? 20 feet? 200 feet? 2,000 feet? How high to see 50 miles around the earth?

Show the math how you get your answer, then I will have a formula!

The formula would include 3 things:

Horizon (what math symbols go here?) How high I am = Distance away the horizon is

Thanks.

Ken

If your happy to accept the earth as being round then http://mathworld.wolfram.com/CircularSegment.html

has all the equations you need to calculate what your after

Al

Thanks for the link. What I am looking for is a formula to find Z if X & Y are known or find Y if Z & X are know. In this diagram if Z=332 feet and X = 0 (right on the water), how far is Y?

http://www.MrChevy.com/distance.jpg

√(h/8 inches)mile

√(72/8=9) miles

at 6'. 3 miles

√(2048/8=256)miles

16 miles

at 170' 8" , 16 miles

I think this should be close enough for anything other than artillery.

RichH

Hello RichH,

You state: Standing on the sea shore, your eyes 6 foot above the ground, looking towards the horizon, the horizon is 3 miles away. What numbers/formula/EQUATION are you using to get that answer?

That is all I am asking for, an equation I can put the numbers in, changing any of them and seeing what the new answer is.

I see you have 6 foot / 8 inches x mile

We don't know the mile(s) and why divide by 8 inches? Could you please make it clearer how you are arriving at your answers (the others you include) and why you use the numbers you do use. How do you get the numbers you use?

My math: eyes @ 6' = horizon is 2.4494 miles away. You say 3 miles, that is a lot of difference in answers.

My math: horizon is 4.8989 miles away = eyes @ 24' above the beach. Plug 4.8989 miles into your equation and how far/high does it say your eyes have to be to see the horizon?

Then throw into the equation changing the height of the horizon itself by looking at the top of a ship, or a mountain, example:

Mt Hood is about 70 miles from where I can see it when driving home. I can see about the top half of it. That is consistant with 4' off the ground in my pickup and it 70 miles away means I can NOT see the lower 5000 feet of it which is hid behind the horizon.

If something (example a tower with a very bright light) were 5 miles high, then we should be able to it 165 miles away, correct? What math is needed to prove that answer?

Thank you,

Ken

i believe he is saying the square-root of (height in inches / 8) = miles

72 inches (6 foot) / 8 = 9

sqrt of 9 = 3 - 3 miles away...

THANK YOU!

Geeez, I sure was stupid not to see that. Thanks for turning the light on!

Ken

Hi, McChevy

My formula is a quick approximation I have used to ballpark height/distance within normal sight range. Your 4.8989 miles comes out to 15.99 ft elevation.

As far as Mt Hood goes, horizon at sea level is based on a zero slope from your position to the horizon. Aroound Mt Hood, I'm sure that you are not at sea level in your PU, and I'd guess that there is a stream or river between you and Mt Hood, meaning that the usual horizon would be well above the land between you and Mt Hood. Try using a spirit level from the side of the road to see where a 0 elevation ( from a level line, same as at horizon) would leave your line of sight.

So an airplane at 40,000 ft with it's landing lights on, would be seen on a beach by a 6'6 person (6' eye height) would, ROUGHLY, 165 miles + 6 miles or 171 miles out.

I used this as a rough guestimator for approximately 1 mile and under. It's accuracy was acceptable in that range. For over 1 mile, perhaps h/11 would be better.

RichH

RichH,

Ahhh, it never dawned on me those weren't check marks but were SQRT. Geeez, talking about going right over my head. All too many times we look but don't see.

My formula was without a factor [ SQRT(HeightOfHriznFt)+SQRT(EyeHeightFoot)=DistanceCanSee ]. How do you get the factor you use [ 8 or 11 ] if I may ask?

So, if I understand what you said (finally!):

√(72/8=9) miles

6 foot x 12 = 72 inches / 8 = 9 and √ of 9 = 3 miles. I can understand that now.

So, airplane @ 40,000 foot x 12 = 480,000 inches / 8 = 60,000 and √ of 60,000 = 244.948974278318 miles. Am I doing it wrong? You say about 171 miles. Where am I going wrong?

Using a factor of 11 is: airplane @ 40,000 foot x 12 = 480,000 inches / 11 = 43,636.3636363637 and √ of 43,636.3636363637 = 208.893187146837 miles. Still off about 40 miles?

Thank you for taking the time to explain this to me. I appreciate it much.

Ken

for your pic illustrate above thats more simple.

assume the highness you stand is h, earth radius is R and the Y you wanted is

(h+R)^2=Y^2 + R^2

h= squrt(Y^2+R^2) - R

vice versa

Y= squrt((h+R)^2 -R62)

where h =Z;

Y= squrt((h+R)^2 -R62)

change at

Y= squrt((h+R)^2 -R^2)

try this formula and to see if it accord to your experience.

OK, I'll give that a try and see what/how it works. I appreciate all the answers I am getting. I will get this figured out yet!

Thank you,

Ken

R²+d² = (R+h)²

So R² + d² = R² + 2Rh + h²

So d² = 2Rh + h²

But the h² term is normally insignificant

So d² = 2Rh

You need to use this twice when "two" objects are involved.

A brilliantly simple explanation, but I always figured the horizon was more than 3 miles away.

-A-

Isn't it though. I guess if you were sitting in a chair on the beach at the water line you could only see about 2 miles out! Some days we learn interesting things... other days we learn things we did NOT want to learn :)

Ken

Thank you Randall, I will poke those formulas into the HP 200LX and see what they work like. I really appreciate you taking the time to reply to my question.

Ken

until h nears or exceeds r

RichH

Im sure there must be some scientific reason that someone will come up with as to why h will never ever come anywhere close to the radius of the earth!!!

Geostationary satelites zip around at about 5r

RichH

How much further out of context from the original question would you like to get?

See answer #21 Bolton Alan, where height is discussed at 18,000 miles. So, yes, h can = much greater than r and still can't see half of the Earth. Like RichH said, no matter how high/far away from the center you travel, you never will be able to see 180º of a sphere.

Ken

Hi:

I had to find this answer as well just recently. Google led me to the Firestick Antenna Tech support site, www.Firestick.com/tech_docs, which had the answers.

For visual line-of sight calcs, Height (ft) divided by 0.5736=X. Then sq. rt of X= distance in statue miles to horizon. Height refers to height-of-eye above sea level (or zero elevation). In the metric system, H(meters)/6.752=X; sq. rt. of X= distance in Kilometers.

For Radio Wave horizon, Let H1= Transmitting antenna height in feet.

Let H2= Receiving antenna height in feet.

Then sq. rt. of H1 x 1.415= D1

And sq. rt. of H2 x 1.415= D2

Summing, D1+D2= radio horizon in statue miles. For the metric system, use meters for height and 4.124 for the conversion constant and the answer will be in kilometers.

Sq. rt. H x 4.124 = Radio Horizon in Kilometers.

Hope this helps

Hank

Hello Hank,

I used to just use √, which would be √26,400'=162.48 miles

RichH says 26,400' (5 miles) / 11 = 28,800' and √ of that = 169.7 miles

You say 26,400' ÷ 0.5736 = 46,025 and √ of that is 214.5 miles?

and then you say v26,400' = 162.48' x 1.415 = 229.91 miles, further yet.

That is a LOT of difference!!

The link doesn't work Hank, could you post a good link that works? I would like to look at the info they furnish.

Thanks,

Ken

Sorry Mr. Chevy:

try this link.

http://www.firestik.com/Tech_Docs/dist2horizon.htm

Hank

Thanks Hank!

That works.

Ken

Considering our planet to be a smooth round Ball with radius R, an observer at height T will see his horizon at distance SQRT(T*(2*R+T)).

As Randell says, if the observer looks at something at height H which is situated behind the horizon the same formula applies.

In other words, if an observer is at height T above ground and he is looking at something behind the horizon at height H, the line of sight distance between them will be SQRT(T*(2*R+T)) + SQRT(H*(2*R+T)).

If our observer wants to know how high should a light be in order to be seen beyond the horizon at a distance D, the formula is:

H = SQRT(R^2+(D-SQRT(T*(2*R+T)))^2)-R. (I hope my parentheses are OK)

This obviously works only where D is beyond the observers horizon.

Enjoy!

Ok, let's see what we have so far Dovy.

Let's use 1' off the ground, which makes multiplying very easy.

SQRT(T*(2*R+T))=Distance away the horizon is

T=1

R=4,000 (roughly)

SQRT(1*(2*R+T))=Distance

SQRT(1*(2*4,000+T))=Distance

SQRT(8,000+1)=Distance

SQRT(8,001)=Distance

SQRT(8,001)=89.45 Miles? No, it should be about 1.22 Miles (at least somewhere between 1 & 2 miles... surely not 89.45 miles).

Would you or someone poke in eyes 1 foot off the sand at oceans edge using your (Dovy's) formula and post the answer please. To make it easy, let's use the earth being 8,000 in diameter so the radius is 4,000 miles.

Randall's post says: So d² = 2Rh

and using 1' again would be SQ(89.44)=8,001 ... [ 8,001=(2*4000*1) ]

That can't be correct...

Ah, a very interesting question I asked. Many different replies, that is for sure!

Everyone's answer should be pretty close to the same, why aren't they I ask.

The one's using the Radius of the Earth don't even seem to be close. Why aren't they? I would think that those answers would be the closest.

RichH uses 8 or 11 as factors when he figures. Can someone show/explain the math of how he arrives at those two numbers instead of 6 or 7 or 9 or 10 or 13 etc.

Thanks for the post Dovy!

Ken

The correct formula should tangent function, because the limit of horizon can only approach 6250 miles (1/4 circumference). Also, except as horizon→0, the h becomes meaningless (the slope approaches 0). So, from 0→90º, the function is soluable, but never reaches 90°. From an infinite distance, you still could not see an entire hemisphere. Close, but never all.

Maybe that will give you guys something to work from. Go radial. The limit is ∏/2 radians.

That's why a quickie, in the head guestimate is better than having to drag out a calculator out on the trail.

RichH

RichH

And it always seems like it is too far if it is late at night and you have that far to go before bedtime.

I appreciate your input RichH, I have always just used √ , which,like you say, works fine out on the road as a guess on distance.

Most (all?) of the other formulas are not put together correctly. Why so many misconceptions I wonder.

Question: Just using √ works pretty good for an estimate. What if the world were 13,000 miles in diameter, could we still use just √ to get an estimate of distance? It doesn't seem so, since the curve would be flatter. So what makes us able to use it with an 8,000 mile diameter Earth?

Ken

With a 13,000 mile diameter, you would have to use a diffent factor, h/16" or so. And depending on the range of possible values.

At 4000 r, the 8" factor was close enough under 2 miles, I think, and the factor of 10 to 12" (Your simple square root √h) is applicable at various ranges to maintain accuracy of any kind. Those little symbols revealed by the blue Ω in the top of the text bar are great, when you get used to them.

RichH

" and using 1' again would be SQ(89.44)=8,001 ... [ 8,001=(2*4000*1) ] "

You're mixing up your units d = SQRT (2*4000 miles *1 foot)

So d = SQRT (2*4000*5280*1)

So d = SQRT (42,240,000) = 6500 feet

Dang.

50% of the people have below average IQ. That is just the way averages work.

Sometimes I feel as if I am at the bottom of that group of people. That was a glaring mistake on my part Randall. Sort of one of those: Duh, how come so stupid today Ken days.

It comes out correct now, Thanks a lot Randall.

I wonder if anyone else is learning from my stumbling around.

Don't feel bad MrChevy. We all make mistakes but we also learn a lot from these posts. I know I have. Its not only very interesting but also a lot of fun to see the many different ways that folks approach a problem or situation.

Cheers,

John

JJ, (I'll bet you never get called that, right? My oldest kid is Johnnie and we call him JB for John Boy most of the time... geeez, 3 more days and he will be 38... how time flys when we are having fun! Gotta remember his birthday since he usually remembers mine.)

I have to feel bad or my jaws would ache from going around grinning all the time. Yep, I make miss steaks (or is that mistakes? LOL), must prove I don't know everything...

Now I remember, I made up a saying years ago: The more I learn, the less I know. /or/ Life is like a ladder, the higher up we go, the more we see, and the more we see, the more we realize how little we know about anything.

Yes, it is good (and fun) to see how others approach problems, this thread has been an enjoyable learning experience for me too.

Have an enjoyable weekend Johnjohn.

Ken

Hi MrChevy,

The formulas I sent you are pure geometry without approximations, therefore the units are consistent. Or:

If you use feet, everything is in feet, when done, scale it up or down as you wish.

If you like algebra you can do the scaling directly in the formulas and get something like 'distance in nautical miles' as a function of 'height in fathoms' or whatever. All this is beyond me, as I am purely metrical. Even my old HP 1000 CX, which is like your HP but without diaries and other gimmicks, is metrical.

I apologize dovy. Dumb me, not paying attention (and I am ALWAYS telling my employees at the shop to PAY ATTENTION)... and then I don't.

You are correct, everything has to be in the correct units to make it work.

Thanks to both you and Randall for bringing that to my attention. It works fine now!

Ken

PS: I have an HP 95LX, 5 of the 200LX's (1 at the shop and the other 4 in different rooms of the house, handy to grab), and a 700LX (the one that took the Nokia phone). I love their SOLVE calculator function with it's "soft keys" so much that I ended up searching for years before finding a Windows program that had that capability. I finally located powerOne®Finance at http://www.infinitysw.com/ for $60 and got the $40 add-on to install it on my Pocket PC PDA. Can make up your own templates to use with it and they have dozens, if not hundreds of ready made templates that are free to download. Some things I miss from the 200LX, but... that is progress, right?

Hi Ken,

There is a company, name forgotten - penalty of old age - who cater for pocket PCs similar to HP 200 LX. If you google for them they'll appear. They repair, upgrade, sell, buy etc.

My 95LX died but the 1000cx is still going strong and it's a very convenient device.

dovy

Hi dovy,

The one you are thinking of is Repairs, Thaddeus Computing, 110 North Court, Fairfield, IA 52556 USA I think. They are the one at http://www.palmtoppaper.com . They sell 95LX's for $135 and 200LX's for $309. I think I bought 7 of the 200LX's off eBay in the past year and never paid over $125 for any of them. Some of them look really close to new. I love eBay!

Old age, ah! I am barely out of my teen years, so I have a long time to go before I have old age problems. (I HOPE). I am NOT looking forward to old age, that is for sure. How does a person know he is getting old anyway? Does someone tell them? LOL

Yes, they are the people whose name I'd forgotten. If you can find an HP 1000 CX on ebay - grab it. It's the 200 LX without any of the installed programs. Purely pocket computer. HP stopped making them for no clear reason.

Dovy's and Randall's formulas are both correct but you need to be consistent with units if you are using 4000 miles as the earth's radius then using T=1 will give you the answer for an elevation of 1 mile. for an elevation of 1 foot you would need to use T=1/5280.

After reading more I see this was already corrected. Sorry for the redundancy

Hi Troy36,

Yep, I see that now, I'm just a little dense sometimes, but after a while it soaks into my brain (I hope).

What do you mean: "Sorry for the redundancy" ...

Without redundancy there would be so little for us to do that we might as well pack our bags and go home (die). Probably 95% of everything that goes on in the world is redundancy Troy, it keeps the world going round. A very small percentage of the 6 billion of us on this ball get to do the new, non-redundant things. The scientists, the engineers, and even some very ordinary folks luck out once in a while and produce something new or a new way of doing something. I have been walking through the door of my automotive machine shop for 29 years... that IS pretty redundant!

Thanks for the post. Have a great weekend.

Ken

Like I said, mine is a rough formula that I don't need to whip out a calculator while I'm driving, to ballpark distances. It's far better than a guess, but not as good as a calc. Using the 1'/mile rather than the 8"/mile, you would have a horizon of 6250 miles from an altitude of 8000 miles. Not too hard to imagine, seeing a full 1/2 of the earth from that height. But in reality, even our geostationary satellites can't quite see a horizon of 6250 miles from their elevation of roughly 18,000 miles.

The 8" figure would yield 18,000 miles for a 6250 mile horizon.

Rough, indeed, but better than a wild guess.

RichH

I am not limited like that.

I can see thousands upon thousands of light years on a clear night...by lookin up...

Yep, if it is a clear night. We get 1.5 of those a year here!

Link from Hank's post:

Excerpt from Firestik Antenna info file:

VISUAL "LINE-OF-SIGHT" CALCULATIONS

In Statute (land) Miles

Height (in feet) divided by 0.5736 = "X"
Square root of "X" = Distance in statute miles to horizon

Example
Height to center of your eye(s) = 5.5ft
5.5ft divided by 0.5736 = 9.588
Square Root of 9.588 = 3.10 statute miles, the maximum distance you could theoretically see if standing on the beach in California looking out across the Pacific Ocean. ^© 2004 Firestik^® Antenna Company

*******************

I wonder how he arrived at the 0.5736 number? Any ideas how he got that number?

MrChevy you are definitely not short sighted.

There are however a modification needed to the formula.

Light changes direction when passing between different densities.

The line of sight is therefore not a straight line. (as seen in mirages)

Where I grew up there were a house just beyond the horizon but this house was actually visible on some days. I think it was on cold days - (when the effect of global warming was felt)

The distance may have been 10 miles with a wide flat valley in between.

Hello Hendrik,

Yes, in certain places on the Earth and at different times of the year... the formula would not be that accurate, but it would still give a person a pretty fair idea how far away something was that was at or near the horizon.

Of course if we whipped out our HP 200LX and entered the formula:

Distance in miles, use factor of 1° = 69.0466 miles

69.0466*acos(sin(hr(LT1))*SIN(hr(LT2))+COS(hr(LT1))*COS(hr(LT2))*COS(hr(LG1)-hr(LG2)))=Distance}

into the SOLVE calculator then we could then just enter LAT & LON and tell how many miles something was away. If I remember correctly I was using the Factor of 69.0466 miles per degree near where I live here at/near the 45th parallel.

Salem, OR 44.57.30.38N 123.00.59.19W Elev 164' Denver, CO Airport 39.51.28.74N 104.40.26.50W Elev 5,341' Poking the distance key gave me: 1,273.20 miles apart using that formula.

Thanks for the post.

Ken

Google says driving it would be 1,307 miles and take about 18 hours 49 minutes. That would be averaging about 69½ MPH. That would be hustling for sure!

I can not add anything usefull in the way of math, but I want to reminisce a little bit. When I was in the U.S. Navy way back in 1966, I can remember being on "watch" on the "flying bridge" at sea. I made 2 atlantic crossings from Norfork, Va. to the Mediterranean sea. While at sea, on a calm day, I remember looking at other ships through the binoculars as they dissapeared over the horizon. At sea, there is a sharp clear line of horizon (if the sea is calm). What is interesting is, after some time has passed, you could see just the tops of the masts of the ships "dancing " up and down! Its a great way to to experience "the curve of the earth" James

Did you receive a response? I am curious what the formula is for determining how far you can see when you are at a 50 foot elevation or 100 foot or 1000 foot. Tony