Original Posted By: "Richard Navratil"
Subject: Re: Pietenpol-List: Fw: [EAA Chapter 25] Navigation problem I need helpwith.I think it's (A). 42 minutes and 84nm.Time to station =Time for Bearing Change(minutes) X60 divided by 10.Time to station = 7X60 = 420/10 = 42.Distance to station = Time to station (minutes) X TAS divided by bearing change. Distance to station = ( 7 minutes) X TAS / 10 = 84.________________________________________________________________________________
Re: Pietenpol-List: Fw: [EAA Chapter 25] Navigation problem I need help
Re: Pietenpol-List: Fw: [EAA Chapter 25] Navigation problem I need help
Original Posted By: "Christian Bobka"
Hi ChrisI'll give it a try.minutes to station=time in seconds divided bynumber of degrees 7x60=420 __________ = 42 mi. 10Dick N.----- Original Message -----
Hi ChrisI'll give it a try.minutes to station=time in seconds divided bynumber of degrees 7x60=420 __________ = 42 mi. 10Dick N.----- Original Message -----
Re: Pietenpol-List: Fw: [EAA Chapter 25] Navigation problem I need help
Original Posted By: "Christian Bobka"
Correction on my last. Should read 42 minutes not mi.Dick----- Original Message -----
Correction on my last. Should read 42 minutes not mi.Dick----- Original Message -----
Re: Pietenpol-List: Fw: [EAA Chapter 25] Navigation problem I need help
Original Posted By: "Christian Bobka"
Sorry, I just re-lookd and noticed the distance part of the problem.distance= seconds divided by degrees 420 7 min x 60 = 420sec _____ = 4242x2mi/min = 84 miles 10deg.Dick N.----- Original Message -----
Sorry, I just re-lookd and noticed the distance part of the problem.distance= seconds divided by degrees 420 7 min x 60 = 420sec _____ = 4242x2mi/min = 84 miles 10deg.Dick N.----- Original Message -----
Re: Pietenpol-List: Fw: [EAA Chapter 25] Navigation problem I need help
Original Posted By: "Richard Navratil"
Tom, Dick, (and Harry too while I am at it),Thanks for the answers. This was a homework problem for Washburn HighSchool. THe instructor sent it to me because he began to doubt his answerwas correct.I think A as well. Just looking at the answers, if you know you are doing 2miles a minute at 120 kts, then the only answer that allows this is A.,chris----- Original Message -----
Tom, Dick, (and Harry too while I am at it),Thanks for the answers. This was a homework problem for Washburn HighSchool. THe instructor sent it to me because he began to doubt his answerwas correct.I think A as well. Just looking at the answers, if you know you are doing 2miles a minute at 120 kts, then the only answer that allows this is A.,chris----- Original Message -----
Re: Pietenpol-List: Fw: [EAA Chapter 25] Navigation problem I need help
Original Posted By: "Christian Bobka"
Christian,It's almost a trick question, it's like test givers pride themselfs onconfusing you with extra information. Example: If a train is heading toBoston at 120 miles and hour, and Corky is adding pinto beans to chilli atthe exact same time, how much does a pound of feathers weigh? Sure you could figure it out with trigonometry, but the only answer thatworks is A based on your true airspeed. And I was all ready to pull out myE6B to correct for crosswinds!Robert HainesDu Quoin, Illinois
Christian,It's almost a trick question, it's like test givers pride themselfs onconfusing you with extra information. Example: If a train is heading toBoston at 120 miles and hour, and Corky is adding pinto beans to chilli atthe exact same time, how much does a pound of feathers weigh? Sure you could figure it out with trigonometry, but the only answer thatworks is A based on your true airspeed. And I was all ready to pull out myE6B to correct for crosswinds!Robert HainesDu Quoin, Illinois
Re: Pietenpol-List: Fw: [EAA Chapter 25] Navigation problem I need help
Original Posted By: "Robert Haines"
OK OK, I'm such a nerd...In 7 minutes at 120knots, the aircraft travels 14NM. This would be theopposite end of the right triangle (remember? opposite, adjacent,hypotenuse) and the distance back to the station off of the 350 radial wouldbe the hypotenuse, since he is traveling 270. Sin of 10 degrees equalsopposite divided by hypotenuse... so after rearranging that equation,hypotenuse equals opposite (14NM) divided by Sin of 10 degrees. Aftertyping all that into my VERY high folutin' HP calculator, rearranging thepens in my shirt pocket, pushing up my thick glasses and blowing my nose,the distance is 80.6NM.... gesh, that's not even a choice!I guess at this point I could back calculate all the distances in theanswers to find the actual distance traveled taking into account acrosswind, since of course all the answers are bigger and the triangle isbigger meaning that more distance was traveled in the 7 minutes meaning theaircraft had a tailwind. But since I don't know the direction or speed itis an infinite amount of combinations to actually equal the distances fromthe station given in the answers... I guess I could iterate through all thepossibilities and find the crosswind directions and speeds that givedistances and times (using the crosswind to correct for groundspeed) thatmatch one (or likely more) of the answers...Meanwhile the student taking the test has just wasted twenty minutes on a 30second question and although he knows the material, the testgiver justscrewed him out of a good grade due to all these damn trick questions.Robert----- Original Message -----
OK OK, I'm such a nerd...In 7 minutes at 120knots, the aircraft travels 14NM. This would be theopposite end of the right triangle (remember? opposite, adjacent,hypotenuse) and the distance back to the station off of the 350 radial wouldbe the hypotenuse, since he is traveling 270. Sin of 10 degrees equalsopposite divided by hypotenuse... so after rearranging that equation,hypotenuse equals opposite (14NM) divided by Sin of 10 degrees. Aftertyping all that into my VERY high folutin' HP calculator, rearranging thepens in my shirt pocket, pushing up my thick glasses and blowing my nose,the distance is 80.6NM.... gesh, that's not even a choice!I guess at this point I could back calculate all the distances in theanswers to find the actual distance traveled taking into account acrosswind, since of course all the answers are bigger and the triangle isbigger meaning that more distance was traveled in the 7 minutes meaning theaircraft had a tailwind. But since I don't know the direction or speed itis an infinite amount of combinations to actually equal the distances fromthe station given in the answers... I guess I could iterate through all thepossibilities and find the crosswind directions and speeds that givedistances and times (using the crosswind to correct for groundspeed) thatmatch one (or likely more) of the answers...Meanwhile the student taking the test has just wasted twenty minutes on a 30second question and although he knows the material, the testgiver justscrewed him out of a good grade due to all these damn trick questions.Robert----- Original Message -----
Re: Pietenpol-List: Apple QuickTime Viewing Problems
Original Posted By: "Michael Fisher"
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