Zflyer Posted January 13, 2021 Share Posted January 13, 2021 This is not a model question but is flight related. I purchased the book 'Flight without Formulae' and it is clearly authored to aid people interested in flight. It poses a number of questions to help in understanding flight and flying. One of the questions : an aeroplane has enough fuel to fly for 4 hours at 100mph. If there is no wind how far can it fly out from base and get home again: that is to say what is its radius of action? Now the answer to that question is 200 miles. The following question is: Will the aeroplane in the above question have the same radius of action if there is a steady wind of 20mph. ( assume you are flying into the wind and have it behind you on the way back.) My initial answer was it would be the same as you gain the wind, however on reflection and re reading the question the answer is clearly determined by fuel use. The answer is given in the book but for the life of me I cannot derive the same answer without some convoluted maths. I will provide the answer from the book following wise readers producing an answer with the relevant explanation. I have another question which caused a hugh row on a radio phone in many years ago. I was in absolute tears of laughter. Quote Link to comment Share on other sites More sharing options...
J D 8 Posted January 13, 2021 Share Posted January 13, 2021 Aircraft flying at 100 mph for 2hr against a 20mph head wind will only get 160 miles from base for a start. But with the availability of a tail wind to assist the return trip it could go further, as to how much the maths are outside my ability. Also an aircraft that runs out of fuel could then glide a fair distance but do not suppose the question allows that. Quote Link to comment Share on other sites More sharing options...
Brian Sweeting 1 Posted January 13, 2021 Share Posted January 13, 2021 I suppose that we need to make at least one assumption which is that the speed given is "ground speed"? Quote Link to comment Share on other sites More sharing options...
john davidson 1 Posted January 13, 2021 Share Posted January 13, 2021 I presume that varying cruising speed on into wind and downwind legs would alter the range for the better, faster upwind and slower down, but the OP question is at a constant speed. If the wind is too great the into wind range could be actually less than 200 miles.Jetsteam wind can reach well above 100mph. Quote Link to comment Share on other sites More sharing options...
kc Posted January 13, 2021 Share Posted January 13, 2021 Model flyers will know that the answer to that question is that the range is less because you need some fuel in reserve and enough to go around again if you cannot land first time. Quote Link to comment Share on other sites More sharing options...
kevin b Posted January 13, 2021 Share Posted January 13, 2021 I cured my problems with complicated theory questions. I left school. I bet Carol what's her name could answer that one. She's a high flyer. Quote Link to comment Share on other sites More sharing options...
Trevor Posted January 13, 2021 Share Posted January 13, 2021 For a distance, D, the time taken for the outbound leg is D/80. For the return leg it’s D/120. The sum of these times is 4hrs. So D/80 + D/120 = 4. Multiply both sides of the equation by 240 to get rid of the fractions and you should be on the home straight, so to speak. Quote Link to comment Share on other sites More sharing options...
Dickw Posted January 13, 2021 Share Posted January 13, 2021 Interesting problem. I am probably wrong, but my take would be that allowing for windspeed you have 80mph ground speed on the way out and 120mph groundspeed on the way back. So:- 80mph for x hours = 120mph for y hours ------- and we know x+y = 4hours So the result of that seems to be a radius of 192 miles. And you say that comes from a book entitled 'Flight without Formulae' ? Dick Quote Link to comment Share on other sites More sharing options...
MattyB Posted January 13, 2021 Share Posted January 13, 2021 Give it 10 mins and the conversation will circle back to this old chestnut... (Before expressing a view I suggest you click the link above and read the article, it's a very good explanation) Edited By MattyB on 13/01/2021 15:41:54 Quote Link to comment Share on other sites More sharing options...
Nigel R Posted January 13, 2021 Share Posted January 13, 2021 is the handbrake still on? Quote Link to comment Share on other sites More sharing options...
Zflyer Posted January 13, 2021 Author Share Posted January 13, 2021 Dick the answer is 192 as you say but telling me adding x+y still doesnt show how i get 192. Maths and sport arent my best subjects. Can you give me the idiots ++ breakdown. Quote Link to comment Share on other sites More sharing options...
Dickw Posted January 13, 2021 Share Posted January 13, 2021 Posted by Zflyer on 13/01/2021 16:48:44: Dick the answer is 192 as you say but telling me adding x+y still doesnt show how i get 192. Maths and sport arent my best subjects. Can you give me the idiots ++ breakdown. Flight out is 80mph for x hours, and that must equal the flight back at 120mph for y hours, i.e. 80x=120y. So: 80x=120y >> x=120y/80 >> x= 1.5y We know x+y=4 hours but we also now know x=1.5y so: 1.5y+y=4 >> 2.5y=4 so: y= 4/2.5 =1.6 hours and therefore x=2.4 hours Back to the first line: 80x=192 and 120y=192 i.e. the journeys out and back. And again, you say that comes from a book 'Flight without Formulae' Dick Quote Link to comment Share on other sites More sharing options...
Gary Manuel Posted January 13, 2021 Share Posted January 13, 2021 Yes 192 miles. D/80 + D/120 = 4 240D/80 + 240D/120 = 4X240 = 960 (multiple both sides by 240) 3D + 2D = 960 (divide the fractions) 5D = 960 D = 192 Is this the answer in the book? Edited By Gary Manuel on 13/01/2021 17:23:21 Quote Link to comment Share on other sites More sharing options...
Martin Harris - Moderator Posted January 13, 2021 Share Posted January 13, 2021 The simplest way to visualise the logic is to consider flying into 100 knot headwind with an airspeed of 100 knots. Downwind you'd fly at 200 mph. Into wind you wouldn't get past the airfield boundary - you'd land with an empty tank having gone nowhere. Flight into wind and back in a more realistic scenario is simply going to give a result somewhere between the extremes of still air when you could travel equal distances and nowhere (into wind) with windspeed matching airspeed. See - no formulae! Edited By Martin Harris - Moderator on 13/01/2021 20:37:29 Quote Link to comment Share on other sites More sharing options...
Phil McCavity Posted January 13, 2021 Share Posted January 13, 2021 Posted by Zflyer on 13/01/2021 13:01:06: This is not a model question but is flight related. I purchased the book 'Flight without Formulae' and it is clearly authored to aid people interested in flight. It poses a number of questions to help in understanding flight and flying. One of the questions : an aeroplane has enough fuel to fly for 4 hours at 100mph. If there is no wind how far can it fly out from base and get home again: that is to say what is its radius of action? Now the answer to that question is 200 miles. The following question is: Will the aeroplane in the above question have the same radius of action if there is a steady wind of 20mph. ( assume you are flying into the wind and have it behind you on the way back.) My initial answer was it would be the same as you gain the wind, however on reflection and re reading the question the answer is clearly determined by fuel use. The answer is given in the book but for the life of me I cannot derive the same answer without some convoluted maths. I will provide the answer from the book following wise readers producing an answer with the relevant explanation. I have another question which caused a hugh row on a radio phone in many years ago. I was in absolute tears of laughter. Without the Maths my answer would initially be No because although the return would be quicker, the fuel burned on the outbound flight will not be made up by the wind assist on the return flight, rather like a car journey along a hilly route would not be equal to the same, downhill, return journey. Quicker but reduced. Edited By Phil McCavity on 13/01/2021 18:09:47 Quote Link to comment Share on other sites More sharing options...
kevin b Posted January 13, 2021 Share Posted January 13, 2021 As far as I can remember the answer is 42. But don't take my word for it. Ask the mice. Quote Link to comment Share on other sites More sharing options...
Zflyer Posted January 13, 2021 Author Share Posted January 13, 2021 Thanks Dick that's a little easier to digest. The book was written by A. C. Kermode, now deceased. I suspect the title lacks a " ? " . In any event it is interesting. Quote Link to comment Share on other sites More sharing options...
Zflyer Posted January 13, 2021 Author Share Posted January 13, 2021 Gary, yes the answer in the book is 192 Quote Link to comment Share on other sites More sharing options...
Guest Posted January 13, 2021 Share Posted January 13, 2021 Well done Dick, Mr. Lane was a good teacher! Quote Link to comment Share on other sites More sharing options...
Peter Jenkins Posted January 13, 2021 Share Posted January 13, 2021 Posted by MattyB on 13/01/2021 15:36:28: Give it 10 mins and the conversation will circle back to this old chestnut... (Before expressing a view I suggest you click the link above and read the article, it's a very good explanation) Edited By MattyB on 13/01/2021 15:41:54 Don't think it did Matty! Still at least the issue of ground speed and air speed got an airing. The OP posed a simple sum, albeit not seen like that by everyone. Best not go looking at the issue of working out how bounday layers work! That gave me a headache many years ago! Quote Link to comment Share on other sites More sharing options...
Doctor Chinnery Posted January 14, 2021 Share Posted January 14, 2021 I reckon Kevin B is right - but you have to make allowances for the Whale and the Petunias as well? However - looking at the problem, we don't seem to be making any allowance for wind resistance - which increases (approximately) with square of the speed of the object through the air. Soooo - your plane is going to use proportionately, significantly more fuel going 'uphill' against the 20mph headwind than when going 'downhill' with the 20mph tailwind. Quote Link to comment Share on other sites More sharing options...
David Holland 2 Posted January 14, 2021 Share Posted January 14, 2021 Doc, the airspeed is exactly the same in both cases. The aeroplane is completely unaware of its groundspeed, it is flying at 100 mph airspeed within an air mass that is moving at 20 mph. Therefore it’s groundspeed is 80 mph into wind and 120 mph downwind but it’s airspeed remains 100 mph and drag remains constant. David Quote Link to comment Share on other sites More sharing options...
Piers Bowlan Posted January 14, 2021 Share Posted January 14, 2021 ZFlyer, this is a point of no return (PNR) question. The formula is:- Endurance X Ground Speed out X Ground Speed back / 2 X True Airspeed so... 4 X 80 X 120 divide by 2 X 100 ANS:- 192, point of no return or radius of action. As David said, the airspeed is constant so no increase of drag - a red herring! BUT, to put the cat among the pigeons. The aircraft will be heavier at the start of its flight, being full of fuel. Therefore more lift will need to be generated at the start of the flight than the end, which mean that more induced drag will be produced too (which is not dependent on airspeed). The consequence is that less power will be required to maintain 100mph at the end of the flight so less fuel consumption. If the fuel flow is not constant this must move the PNR further from the origin? Discuss! 'Flight Without Formula' A brilliant book for anyone interested in aircraft flight Edited By Piers Bowlan on 14/01/2021 10:06:34 Quote Link to comment Share on other sites More sharing options...
Dickw Posted January 14, 2021 Share Posted January 14, 2021 Posted by Peter M on 13/01/2021 23:40:21: Well done Dick, Mr. Lane was a good teacher! Wow! That jerked some memories! Must be nearly 60 years ago - but you have an advantage over me as I didn't keep in touch with anyone or keep any records from that period. I assume we knew each other back then? Dick ps ??? Did you ever own a Morgan? Edited By Dickw on 14/01/2021 10:19:51 Quote Link to comment Share on other sites More sharing options...
Chris Walby Posted January 14, 2021 Share Posted January 14, 2021 but what happens if the aircraft is less efficient at the lower mass without retrimming? will it have less or more drag? IIRC gliders sometimes carry water as ballast to suit certain weather conditions and beat others without ballast over a measured course/time? How complex can we go without formula Edited By Chris Walby on 14/01/2021 11:17:17 Quote Link to comment Share on other sites More sharing options...
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