tag:blogger.com,1999:blog-10000144.post111117848402445258..comments2024-03-13T09:47:40.487+00:00Comments on Cockpit Conversation: A Third Post about Timing HoldsAviatrixhttp://www.blogger.com/profile/13634111275860140084noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-10000144.post-20064483698198737092008-06-05T14:54:00.000+00:002008-06-05T14:54:00.000+00:00The inbound is supposed to take one minute.The inbound is supposed to take one minute.Aviatrixhttps://www.blogger.com/profile/13634111275860140084noreply@blogger.comtag:blogger.com,1999:blog-10000144.post-7564547869193525132008-06-05T11:46:00.000+00:002008-06-05T11:46:00.000+00:00I was taught to not worry about the actual time of...I was taught to not worry about the actual time of the inbound leg and just add whatever the headwind component is to 60 to give me the time in seconds to track outbound. <BR/><BR/>However if you wanted to be precisely correct you could do what I have detailed below (with explanations). Trying to work it out in the air while maintaining straight and level on the inbound would be a challenge though. (The working and notes were more for my benefit as opposed to yours btw.)<BR/><BR/>Let:~<BR/>GS(I) = inbound groundspeed<BR/>GS(O) = outbound groundspeed<BR/>w = wind component in the direction of the beacon parallel to the inbound leg of the hold<BR/><BR/>So:~<BR/>GS(I) = TAS + w<BR/>GS(O) = TAS - w<BR/><BR/>You said you want the inbound leg to be exactly one minute from when you roll wings level at the end of the turn inbound.<BR/><BR/>Therefore let's call the distance inbound d(I) and we can say that:~<BR/>d(I) = st <BR/>= GS(I) × 1/60<BR/>= GS(I)/60<BR/><BR/>Assuming that there is a tailwind on the inbound part of the hold you will have already rolled wings level before getting abeam of the beacon to head outbound due to the wind pushing you further inbound in the turn. So the timing will start from when you are abeam the beacon. We know that the distance travelled has to be the same as on the inbound plus a little extra. <BR/><BR/>Let us break this down into two parts. One part is you travelling to a point abeam where you would roll out of your inbound turn to do your one minute inbound. I am going to call this distance outbound one or d(O1). However if you start turning here, due to the wind, we know that you will end up too close to the beacon and your inbound leg will take less than one minute. So you need to fly a little further and I will call this distance required distance outbound two or d(O2).<BR/><BR/>We know that d(O1) = d(I) so we can work out the time taken to fly this portion of the outbound. Let's call the time required t(O1):~<BR/>t(O1) = d/s <BR/>= d(O1)/GS(O) <BR/>= d(I)/GS(O) <BR/>= GS(I)/60 / GS(O) (since d(I) = GS(I)/60)<BR/>= GS(I) / 60GS(O) (rearranging)<BR/><BR/>The little extra distance travelled needs to be just enough to allow for the wind pushing us towards the beacon as we turn inbound. We know that the turn takes one minute so we can figure out how much you will get pushed by the wind.<BR/><BR/>So:~<BR/>d(O2) = st <BR/>= w × 1/60 <BR/>= w/60<BR/><BR/>Now we can work out the extra time needed to fly outbound, which we will call t(O2):~<BR/>t(O2) = d/s <BR/>= w/60 / GS(O)<BR/>= w / 60GS(O) (rearranging)<BR/><BR/>So the total outbound time, t(O):~<BR/>= t(O1) + t(O2) <BR/>= GS(I) / 60GS(O) + w / 60GS(O)<BR/>= (GS(I) + w) / 60GS(O)<BR/><BR/>But currently our speeds are in distances per hour so our time will be in hours. Therefore the total time in minutes which is what we want is 60 times this and:~<BR/>= (GS(I) + w) / GS(O)<BR/><BR/>Or in English your time outbound is the ratio of your inbound groundspeed plus the component of wind acting inbound all over your outbound groundspeed.<BR/><BR/>Alternatively we could substitute back in the equations with our TASs in which case:~ <BR/>t(O) = (TAS + w + w) / (TAS - w)<BR/> = (TAS + 2w) / (TAS - w)<BR/><BR/>This will only work if you have a tailwind on the inbound leg. If you have a headwind inbound you have the complication of already being past the abeam point when you start your outbound timing which will screw things up. I'm sure it would be possible to figure out but my brain hurts right now so maybe some other time.<BR/><BR/>Interestingly enough though, as I mentioned to start with, the rule of thumb I was taught for your outbound timing is that you should travel outbound for 60 seconds plus one second per knot of headwind. I am learning to check rules of thumb though as they have been wrong in the past. This one is no exception as I have just done the maths (which is quite long but relatively easy in this case) and this rule only gives you a one minute inbound if your TAS happens to be 180 knots plus the wind (or the wind happens to be your TAS minus 180 knots ). The plane I am training in has a holding speed of 120 knots so the rule we have been taught will not work precisely, although I am sure the difference is probably so slight as to be inconsequential.<BR/><BR/>As an aside though, is the inbound supposed to take one minute or is one lap of the hold supposed to take four minutes?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-10000144.post-1166426751402627872006-12-18T07:25:00.000+00:002006-12-18T07:25:00.000+00:00These 'Holding Timing' posts make my head hurt! H...These 'Holding Timing' posts make my head hurt! How happy am I for FMS that figures these things out and just leaves me alone!Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-10000144.post-1111253618293722102005-03-19T17:33:00.000+00:002005-03-19T17:33:00.000+00:00They usually do things better in Europe, except fo...They usually do things better in Europe, except for the UK, where we seem inexplicably bound to American practice in most thingsAnonymousnoreply@blogger.com