Just One Degree.....

[Garmin8888]

If one is trevelling in a straight line and one goes of the original straight line course by just one degree how far apart whould one travel in say 10 miles........... (the differance)?

[+stu_and_sarah]

If one is trevelling in a straight line and one goes of the original straight line course by just one degree how far apart whould one travel in say 10 miles........... (the differance)?

Halve the angle, take the sine of it. Multiply by the distance you walked. Multiply the result by 2.

 

So for 10 miles at one degree, that's 0.17 miles (898 feet) out.

 

Cheers,

 

Stu

[NickPick]

The problem is a right angle triangle. Your distance walked will be the hypotenuse (longest side), and the crosstrack error (distance from where you should be) is the opposite side.

 

Pythagoras tells us that the sine of the angle equals the opposite divided by the hypotenuse. Re-arrange this to make:

 

the opposite side = hypotenuse x sin of the angle

 

crosstrack = 10miles x sin(1 degree)

=10 x 0.0174

 

= 0.174 miles

 

or about 280 metres, which is a bit too far away to start looking for the cache!

 

This is why it's important, when walking / sailing on a bearing and distance, to re-establish your position by landmarks when possible.

[Garmin8888]

Thanks u 2

 

I always mark the tall landmarks where I go walking, ---- as when one is in the fields one can see the direction of them...........

[+stu_and_sarah]

The problem is a right angle triangle.

I initially thought that, but it's not quite true.

 

It's actually an icoceles triangle which can be modelled as two right triangles. The right-angle method will give the error from your original track but NOT the error from the actual intended destination.

 

Try it for large angles (eg. 179 degrees off track) and you'll see the difference.

 

By the incorrect method, 10 miles at 179 degrees off track puts you 0.17 miles from your TRACK.

 

By the icoceles method, 10 miles at 179 degrees off track puts you 19.999 miles from your TARGET. This is intuitive, since you've just walked 10 miles almost directly away from the target which was 10 miles away to start with.

 

Cheers,

 

Stu

[NickPick]

Of course it is. I was thinking too simply this morning! With an angle of 1 degree, the rounding resolution makes the difference negligible, it becomes obvious when we get to larger angles.

 

Thanks for the explanation,

Nick

[+stu_and_sarah]

Of course it is.  I was thinking too simply this morning!

And for my part, I just made the error of saying 'icoceles' triangle, when I meant 'equilateral'.

 

And managed to spell it 'icoceles' instead of 'icosceles'.

 

And now I've started three sentences with 'And'

 

Oops!

 

Cheers,

 

Stu

Edited May 17, 2005 by stu_and_sarah

[+stu_and_sarah]

Oh well... in for a penny.

 

I did mean 'icosceles'.

 

I'll quit now.

 

Stu

- If you're going to mess up - do it with style.

[NickPick]

If your angle was 60 degrees, it would be equilateral!

[+Simply Paul]

I hate to be obtuse, but- No Paul, walk away, walk away...

 

SP

[+Bill D (wwh)]

isosceles

[+The Forester]

Pilots have to do this calc all the time to work out wind drift angles. In basic nav training they are taught a useful shortcut which enables swift mental arithmetic to be used.

 

It's called the 1 in 60 rule. If you steer 1° off course you will be 1 mile offtrack after 60 miles. After 10 miles you will be a sixth of a mile offtrack.

[Garmin8888]

wow only 60 miles .......?

[+The Forester]

Old joke:

 

Navigator to Pilot: "Turn 1° to Starboard"

 

Pilot to Navigator: "Don't be silly. I can't make a turn of only one degree"

 

Navigator to Pilot: "Turn 2 degrees to Port"

 

Pilot to Navigator: "That's better. I can make that turn"

 

Navigator to Pilot: "Now turn 3 degrees to Starboard".

[Jumbo Village]

Is that a joke? If so, then can I tell an equally hilarious one?

 

A guy walks in to a pub and notices that there are two pieces of meat stuck on the ceiling. So he asks the barman what's going on:

 

"Ah, well, it's a competition you see." he says, "You put a pound in this jar, and if you can reach the meat, you win the money in the jar and the meat... Fancy a go?"

 

The man thinks about it for a while and then replies:

 

"No thanks, the steaks are too high."

 

Hurrah!

[+Deneye]

Doh!

 

I like the 1-in-60 rule of thumb but i'd say it's only good for shorter distances as there is a bit of error involved. Although with it being an error-type (1° off course) calculation, maybe they'll cancel each other out and you'll be spot-on target

 

is that 1° to the left or to the right? Why would you halve the angle as mentioned in earlier posts? You'd be better off just taking the tangent of the angle, multiply by distance and then doubling the product.

[NickPick]

Tan of the angle will give you the same value as dividing the distance from the intended track by the distance made good along the track.

 

Using such a narrow angle as 1 degree, using tan is virtually the same as using my original answer of distance x sin(angle). This actually gives you the distance away from the intended track, not the distance away from the target. In my example, the answer was 0.17452 km. This is actually the distance from a point 9.9984km along the track.

 

However, as Stu pointed out, if we increase the angle (I'm going to increase it to 20 degrees), then using my formula will put us 3.42 km away from a point 9.396 km along the track. (check the root of the sum of these values squared and it gives us 10km travelled). so we're 3.42km away from a point on the track 0.603km from the target. Pythagorus calcs now give us a distance from the target of 3.472km

 

Using Stu's method, we take the intended track, and the actual course, both straight lines at an angle of 20 degrees. This makes an isosceles triangle. to calculate the other side of an isosceles trinagle, we have to split it in half to make 2 right triangles, then use trigonometry to calculate the length of the opposite side (sin = opposite/hypotenuse), then double it to give us the final distance from target.

 

The ultimate test of any method is to try this: If we were 180 degrees off our bearing (ever followed the blue end of a compass instead of the red?) and travelled 10km, then my method would give us a distance away from the line of the intended track (which would be zero). Stu's method would give us 20km.

 

It's ever so difficult to explain this without drawing it, so if anyone wants a drawing, let me know and I'll do one some time.

[+stu_and_sarah]

It's ever so difficult to explain this without drawing it, so if anyone wants a drawing, let me know and I'll do one some time.

Indeed... I have a bunch of scrap papers on my desk now covered in triangles and trig formulae amongst my work notes.

 

Hopefully, I won't accidentally mix them up. If you suddenly see your tv picture turn 1 degree and move 10 miles away, you'll know why.

 

Stu

Edited May 19, 2005 by stu_and_sarah

[+Wuthered]

Halve the angle, take the sine of it. Multiply by the distance you walked. Multiply the result by 2.

 

 

And for small angles sin(theta)=theta (in radians)

So

Halve the angle :- pi/360

Take the sine of it :- pi/360

Multiply by 10 miles :- pi/36

Multiply by 2 :- pi/18

 

=0.1745 miles

 

So Foresters 'rule of thumb' approximates pi to be 3.

 

Funny, my tv has just put on an anorak and walked out.

[+Tupperware Hunters]

my brain hurts

[Garmin8888]

So glad that I bought an a Garmin lol