# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

**Re: Regarding fix bearings**

**From:**Greg R_

**Date:**2008 Dec 21, 10:31 -0800

--- On Sun, 12/21/08, Gary J. LaPookwrote: > Another method is taught to every instrument rated pilot. > With the station near the wing tip (near the beam if talking about > ships) measure the time it takes for a certain number of degrees of > bearing change in seconds. Then divide the seconds by the number of > degrees of bearing change and the result is the distance to the station > in minutes. Seems like I remember reading about that method in my instrument ground school books way back when (mid-70s), but I can't recall ever using it in practice (we mostly used DME back then, and GPS these days). I think you got some/most/all of your flight training in the military (?), so that curriculum might have been different from what us civilians were taught. Of course, alternate methods of navigation are always good to have in your back pocket, and I wouldn't sell it short by any means. -- GregR --- On Sun, 12/21/08, Gary J. LaPook wrote: > From: Gary J. LaPook > Subject: [NavList 6784] Re: Regarding fix bearings > To: NavList@fer3.com, "Beverley Maxwell" > Date: Sunday, December 21, 2008, 1:59 AM > It looks like the classic "double the angle on the > bow" technique. I am > attaching Dutton's explanation. > > Another method is taught to every instrument rated pilot. > With the > station near the wing tip (near the beam if talking about > ships) measure > the time it takes for a certain number of degrees of > bearing change in > seconds. Then divide the seconds by the number of degrees > of bearing > change and the result is the distance to the station in > minutes. This > works because the tangent of one degree is approximately > equal to 1/60 > and this approximation holds true for small angles. > The actual formula is (time/degrees of bearing change) x 60 > = time to > the station. By measuring the time in seconds the > multiplication by 60 > is accomplished automatically when you take the result in > minutes. The > formula can also be worked to find distance. Simply measure > the time in > minutes and multiply by your speed in knot, then divide by > the number of > degrees of bearing change to give you your distance off in > nautical > miles. Measuring in minutes and multiplying by your speed > takes care of > the internal multiplying by 60. In flying the bearings are > measured > with and ADF or VOR but the same technique works with > visual bearings. > > A simple example. Say it takes 120 seconds for the bearing > to change 10 > degrees. 120/10 = 12 minutes to the station. The same > example, 120 knots > times 2 minutes divided by 10 puts you 24 NM from the > station. This is > the same result as multiplying the 120 knot speed by 12 > minutes divided > by 60 minutes per hour. I have attached an excerpt from FM > 1-240 that > illustrates this. > > gl > > George Huxtable wrote: > > inuik@yahoo.com, or A-08, wrote the posting copied > below. > > > > That idea sounds original, simple, and interesting. > > > > Tell us more, A-08. > > > > George. > > > > contact George Huxtable, at george@hux.me.uk > > or at +44 1865 820222 (from UK, 01865 820222) > > or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 > 5HX, UK. > > > > > > ----- Original Message ----- > > From: > > To: > > Sent: Saturday, December 20, 2008 2:12 AM > > Subject: [NavList 6773] Regarding fix bearings > > > > > > | > > | Regarding fix bearings: A convenient practice of > taking two bearings at > > one landmark is to take the first bearing at 45 deg, > then second bearing is > > taken off beam (90 deg). In this case distance off > beam is equal distance > > traveled. > > | There are also few so-called "magic" > angles which play the same trick: > > upon using them distance travelled is equal distance > off. > > | > > | I wrote a formula which permits for the first > bearing chose any angle you > > want and then calculate the angle for the second > bearing which, as a result, > > equals distance travelled to distance off. > > | > > | I think it gives some flexibility in bearing taking > and especially in > > calculating distance off. > > | > > | Respectfully, > > | > > | A-08 > > | > > | > > | ------------------------------------------ > > | [Sent from archive by: inuik-AT-yahoo.com] > > > > > > > > > > > > > > --~--~---------~--~----~------------~-------~--~----~ Navigation List archive: www.fer3.com/arc To post, email NavList@fer3.com To unsubscribe, email NavList-unsubscribe@fer3.com -~----------~----~----~----~------~----~------~--~---