A330300 From United States of America, joined Nov 2005, 174 posts, RR: 0 Posted (15 years 2 months 3 days 6 hours ago) and read 5479 times:

A contoversial problem NO ONE in my class of 30 people got right. (Of course, that depends on how you see the problem)

A jet is flying from Honolulu to San Francisco, a distance of 2400 miles. Traveling through still air, the jet travels at 600 mph. It has a 40 mph tailwind. If the jet has an emergency, after how many hours would it be faster to go on to San Francisco than to turn back to Honolulu?

--------------My logic below-------------

When I did the problem, I assumed that the aircraft was traveling at 600 mph East, and 520 mph west. There are two ways to solve the problem, one with a distance/rate+time ratio and a guess and check way. I guessed and checked and arrived at an answer of 1.853333333... hours or 1/ 113/120 hours. Of course, that was "wrong." It seems like both the book and my "teacher" (who really is just someone who rewrites examples from the book and relies on the answer key heavily) agree that the aircraft is traveling at 640 mph. This could work, but the wording of the problem is tricky. Even more shocking, my teacher has trouble grasping the concept that a tailwind one way turns into a headwind the other way....so at first, she simply divided the distance by speed to arrive at her answer. I quickly alerted her to the wording of the question, and showed her how I arrived at the answer of 2 hours, if we do in fact use 640/560. she is now "discussing the question with other teachers." My question for HTers is....what is your understanding of the problem? Should I get credit for my "wrong" answer? Even though this problem is highly unrealistic and does not incorporate logistical factors, what should the correct answer be?

N766AS From , joined Dec 1969, posts, RR:
Reply 1, posted (15 years 2 months 3 days 5 hours ago) and read 5368 times:

I agree completely with you. Many math/science text books dealing with aviation-related questions are inaccurate. Here is one that I have come across:

"An airplane passing over Richmond at an elevation of 33,000 feet begins its descent to land at Washington, D.C., 107 miles away. How many feet should the aircraft descend per mile to land in Washington, which has an elevation of 25 feet?"

Number one: Aircraft don't descend in feet-per-mile. They descend in feet-per-minute (and actually using the correct terms may make the problem a bit more challenging).

Number two: I am sure they aren't teaching the descent formula...

Iainhol From , joined Dec 1969, posts, RR:
Reply 2, posted (15 years 2 months 3 days 5 hours ago) and read 5362 times:

The problem states the aircraft travels through still air at 600 mph. But the problem is written very badly stating the aircraft always has a tail wind. So basically it is 1200 miles or 1.875 hours.
>>my teacher has trouble grasping the concept that a tailwind one way turns into a headwind the other way<<
That is not what the question said, it claims the plane has a tailwind the whole way which blows at 40 mph.
You would have a vey good arguement with that answer but the answer they are looking for is.

You are almost there I got 1.906 hours and a distance of 1220. But after that it would take 2 hours to get to San Fransico with a distance of 1280. You need to divide the distances by the speed 640 mph, with out that it is just time to get back which is 2 hours eitehr way. You should get some credit but you did miss something out which left 6 minutes off!!
Iain

Iainhol From , joined Dec 1969, posts, RR:
Reply 3, posted (15 years 2 months 3 days 5 hours ago) and read 5350 times:

"An airplane passing over Richmond at an elevation of 33,000 feet begins its descent to land at Washington, D.C., 107 miles away. How many feet should the aircraft descend per mile to land in Washington, which has an elevation of 25 feet?"

Feet per minute would be the next step, and we would also need a speed!
Here is the answer: 308.17 feet per mile.

If the plane travelled at an average speed of 325mph its decent rate would be: 1735.52 feet per mintues.
Iain

Charles802 From United States of America, joined Jul 1999, 380 posts, RR: 0
Reply 4, posted (15 years 2 months 3 days 5 hours ago) and read 5342 times:

No correct answers so far, I don't think

Anyway, Iainhol... 1220+1280=2500
The distance is only 2400. You would not want to be more than halfway to SF, and then turn back.
If the plane flew from Hawaii to SF, it would take 3 hours and 45 minutes (Not accurate, of course).
It would be before 1 hour and 52.5 minutes.
The speed should be 640 going east, and 560 going west.

I don't have time to do the math right now, but I'll try later.

Charles802 From United States of America, joined Jul 1999, 380 posts, RR: 0
Reply 6, posted (15 years 2 months 3 days 4 hours ago) and read 5327 times:

Actually, I think Iainhol is right. But, it should have said 1120+1280=2400. So, after 1120 miles you should continue to SF.

Point 1 is the equal time point. 1120nms (1hr 45mins.)
Point 2 is the equal distant point in still air. 1200nms
Point 3 is the distance travelled eastbound after 2 hours. 1280nms.

Point 1 and 3 assume 40 mph wind(tail and headwind respectively) and an elapsed trip time of 3 hrs and 45mins.
Therefore the time and distance where it would be quicker to return to HNL than continue on to SFO(assuming no other factors ie. descent which results in a slower airspeed) would be 1119nms or approx. 1hr. and 44.9mins.
Hope this helps!

A330300 From United States of America, joined Nov 2005, 174 posts, RR: 0
Reply 9, posted (15 years 2 months 3 days 4 hours ago) and read 5315 times:

Thanks for all the replies- I believe Gate Keeper is nearest to the correct answer, which another message board's poster explained to me-

The question is asking at what point it is faster to continue to San Francisco then to return to Honolulu. So we should set the equation where it takes same time to continue to San Francisco and to return to Honolulu, mid point in terms of time not distance.
Let say that the mid point in time is X miles from Honolulu. And the mid point in time is Y miles from San Francisco. Here we can set up the equation X + Y = 2400.
Since it is the mid point in time so it will take same time to continue to San Francisco as well as return to Honolulu. There is a tail wind so if we continue to San Francisco, we will be traveling in 640 mph. If we return to Honolulu from that point then we will be traveling 560 mph. Here we can establish the equation 640/Y=560/X
If we solve above simultaneous equations, X=1120 miles. Which the mid point in terms of time is 1120 miles from Honolulu. If we are traveling in 640 mph, we will reach that point in 1 hour and 45 minute.
X + Y = 2400
640/Y = 560/X
From the first equation we can derive Y = 2400 – X. Substitute that into second equation will give you:
640/(2400-X) = 560/X
rearranging the above equation will give you
640X = 560 (2400 – X)
when
640X = 1,344,000 – 560X
again
1200X = 1,344,000
Finally
X = 1120, which is 1120 miles from Honolulu.
So answer is after 1 hour and 45 minute of flight it is faster to continue to San Francisco then to return to Honolulu.

N863DA From United States of America, joined Sep 2004, 48 posts, RR: 5
Reply 10, posted (15 years 2 months 3 days 4 hours ago) and read 5311 times:

Hi Y'all

In an aviaiton math textbook for intermediate math (Hey I didn't do it for four years and I'm real bad at it) at college, one question is:

"According to figures provided by the ATA of America, the Boeing 767-400 and McDonnell Douglas L-1011-200/-400 are amon the air carriers with the maximum passenger seating. The Boeing seats 110 more passengers that the McDonnell Douglas, and together the two models seat 696 passengers. What is the seating capacity of each type?"

It works out at B767-400: 403 (which, incidentally, just happens to be right for a 767-400!!! but at the time the book was written the plane didn't exist, and the 747-400 holds a lot more people) and the MDC L-1011 (spot the mistakes) 293. (hardly maximum stats for the airliners.)

Matt d From United States of America, joined Nov 1999, 9502 posts, RR: 43
Reply 11, posted (15 years 2 months 3 days 4 hours ago) and read 5313 times:

A very good question indeed, but did anyone factor in that a plane is also SLOWING DOWN while decending? A plane may be travelling at 450 MPH (or knots) while at FL330, but by the time it reaches 1000 AGL, it might only be travelling around 170 mph.
Was that taken into account or does that just further complicate matters?

Gate Keeper From Canada, joined Jan 2000, 176 posts, RR: 0
Reply 14, posted (15 years 2 months 3 days 3 hours ago) and read 5289 times:

As a pilot I am fortunate enough to fly around the world and this question posed is something an ETOPS(engines turn or passengers swim) pilot faces virtually every flight. Normally the ETP(equal time point) is calculated using an average ground speed which includes slower speeds during climb and descent. This ETP is depicted on the flight plan in 3 ways. Latitude and Longitude, distance(nms) and time in hours and minutes.

FlyBoeing From United States of America, joined May 2000, 866 posts, RR: 2
Reply 15, posted (15 years 2 months 3 days ago) and read 5240 times:

A330300

The simple solution is to set both "times to return" to the same:

Time to return if returning to HNL after t hours of flight

(2400 - t(640))/640

Time to return if returning to SFO after t hours of flight

t(640)/560

Then you're golden. Just solve for t

I feel sorry that your teachers can't even get the simplest answer from the solutions manual. I did that in fifteen minutes. But it's good that you questioned the teacher, because letting the blind lead the blind only gets you into trouble. The only problem is that you guessed and checked. Don't. It's never a valid way to get to a mathematical answer. Could you be responsible to your passengers if you got on the intercom and said "Ladies and gentlemen, by guessing and checking, I figured out that we'd better turn back"

Iainhol From , joined Dec 1969, posts, RR:
Reply 16, posted (15 years 2 months 2 days 23 hours ago) and read 5227 times:

Fly Boeing,
I agree about guessing and checking, but you can get a formula out of it which can help out! Physics seems to be the hardest class for teachers to teach, my physics teacher was OK, he went over answers in class which really did not help, but if you did have questions he could answer them, and would happily stay after class and help you out. But all test where open book, notes, and homework, which prevented most of us from studying!
Iain

Hypermike From United States of America, joined Dec 1999, 1001 posts, RR: 5
Reply 18, posted (15 years 2 months 2 days 12 hours ago) and read 5193 times:

Sheesh... And I was having trouble grasping the whole hydroplaning velocity based on tire pressure formula.