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Anonymous Poster

Algebra 2 Question!!!

11/12/2007 5:42 PM

Okay, I'm a high schooler, and I get extra credit if I do this cuz the teacher can't. :]

Physic Application:

An Airplane is flying due west at 400 mph when the wind is blowing south at 30 mph. What is the speed of the plane with no wind resistance? Use the pythagorean theorem to draw and label a diagram.

I have no idea what's going on. How do you calculate perpendicular wind resistance, and how does pythagoras' theorem figure into this?

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#1

Re: Algebra 2 Question!!!

11/12/2007 5:50 PM

Think of it as a canoe paddling across a river. Where does the canoe need to point and how fast should it paddle to land at a spot on the bank directly across from the start point?

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#2
In reply to #1

Re: Algebra 2 Question!!!

11/12/2007 6:08 PM

BTW - this is an excellent intro to vector algebra. If this is the type of thing that turns your crank, you just may be an engineer in the hatching!

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#3

Re: Algebra 2 Question!!!

11/12/2007 6:45 PM

If it was a slow old biplane doing 40mph due west (ground-speed) with a 30mph wind blowing south, or maybe a plane doing 120mph with a 50mph wind, do these numbers suggest anything to do with Pythagoras' theorem which may help?

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#4

Re: Algebra 2 Question!!!

11/12/2007 10:44 PM

We aren't in the habit of doing somebody's homework.

Spend some energy. Look up pythagorean theorem. Apply the labels and do the calculations!

TANSTAAFL

(You can google that and get the meaning.)

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#5
In reply to #4

Re: Algebra 2 Question!!!

11/12/2007 10:49 PM

It's a poor statement of our education system that the teacher is not able to figure this one out.

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#6
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Re: Algebra 2 Question!!!

11/12/2007 10:52 PM

Agreed, but we don't know what country this person is from. (most likely the US but????)

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#7
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Re: Algebra 2 Question!!!

11/12/2007 10:59 PM

True.

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#14
In reply to #5

Re: Algebra 2 Question!!!

11/14/2007 1:46 AM

... in 6th grade, let alone algebra 2!

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#38
In reply to #5

Re: Algebra 2 Question!!!

11/14/2007 4:11 PM

Hi CSM,

I suspect the teacher is quite capable of "figuring the problem out", but is forcing the student to actually do the work. Guest said "can't" but in reality he probably means "won't".

Guest said "Okay, I'm a high schooler, and I get extra credit if I do this cuz the teacher can't. :]"

Notice Guest did not say the teacher "can't solve the problem" but he/she just said "teacher can't. :]". To me that says the teacher "can't tell him/her the answer because it up the him/her to do the work.

-John

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#39
In reply to #38

Re: Algebra 2 Question!!!

11/14/2007 4:40 PM

Hope springs eternal! Although with as much discussion and disagreement as we've seen on this thread of (presumably) educated contributors, maybe it really is to much to expect a H.S. teacher to be able to answer what I thought was a pretty straight-forward problem.

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#40
In reply to #38

Re: Algebra 2 Question!!!

11/14/2007 4:46 PM

I agree, we don't deal with the method's it takes to encourage teens, in a classroom setting day in and day out. I get the feeling this was a ploy to provide enthusiasm.

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#8

Re: Algebra 2 Question!!!

11/13/2007 3:53 AM

"The square of the hypotenuse is equal to the sum of the squares of the opposite two sides."

There's an ingenious proof based upon two squares, one smaller than the other and inside it, touching the larger square at the four corners of the smaller square. It's doodle time!

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#9

Re: Algebra 2 Question!!!

11/13/2007 4:40 AM
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#10

Re: Algebra 2 Question!!!

11/13/2007 12:51 PM

That's easy. 400MPH is the correct answer.

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#11

Re: Algebra 2 Question!!!

11/13/2007 5:09 PM

Is the 400 and 30 correct?

I would say a speed of 400 with a wind of 300 would have been a nice choice =

3*3 + 4*4 = 5*5 = A air speed of 500km

Maybe this question was adapted from an horse cart example.

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#12
In reply to #11

Re: Algebra 2 Question!!!

11/13/2007 7:45 PM

Looks like another case of Guest, here. We'll probably never find out any more about this non-contributor, who has taken up our time trying to give a reasonable answer.

(BTW I'm replying here to all the folk who did something useful in response to the original post).

This is the 3rd thread I've looked at tonight to which I've contributed - and the 3rd time I've tried to express how pissed off I'm beginning to get, trying to help, without any thanks, or even any feedback. I gave up with my comments each time, because it sounded too much like just another whinge.

What concerns me most is that I'm beginning to become inured to the recurring scenario, to the point where I let loose at any hapless Guest (or Browser), lately to the point where I end up with egg on my face.

Can anyone thing of a way out?

(Sorry to rant in a reply to you, Pussycat - it certainly wasn't directed at you).

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#18
In reply to #12

Re: Algebra 2 Question!!!

11/14/2007 8:04 AM

Maybe it is time to stop answering guests questions, its easy enough to sign up. Quite a lot of the questions could be Googled any way.

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#13

Re: Algebra 2 Question!!!

11/13/2007 10:32 PM

QUOTE: What is the speed of the plane with no wind resistance?

IF there is NO wind resistance, then it doesn't matter what speed the wind is blowing or from what direction. It has NO effect on the planes speed.

If the wind were blowing due east at 400MPH it would make no difference on the planes speed. NO wind resistance.

An IQ type question, IF that is the way the question WAS written by the teacher.

A sailboat is out on the lake and the wind is blowing past it at 50MPH, but there is NO wind resistance... how fast is the sailboat moving? Zero MPH from the wind, that is for sure.

Read what the question says, not what you think it says.

Look at that building, it is GREEN... is it? Have you been all the way around it? Or are you assuming? Correctly would be: the side I can see is GREEN, don't know what the color of the other sides are. At one time, years ago, you could walk into the front hall a couple steps and be able to see 7 different walls, painted 4 different colors.

It's all in the details.

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#65
In reply to #13

Re: Algebra 2 Question!!!

11/29/2007 8:13 AM

Correct, Mr. Chevy! I don't think most posters have read the question carefully. It says if there were no wind resistance, not if there were no wind. Perhaps it's just badly worded and he meant the same, but maybe it's a trick question. It needs clarifying.

If wind resistance disappeared (but lift maintained, presumably) plane would go faster but how much depends on type of propulsion etc, and it's a bit of a hypothetical question. Also if there's no wind resistance presumably the wind blowing south has no effect. (I understand what Blink says - wind action isn't a matter of resistance, it's that the plane is flying thru a moving body of air, that's why it's a hypothetical question)

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#15

Re: Algebra 2 Question!!!

11/14/2007 2:12 AM

If there is no wind resistance a plane cannot fly.

This is a standard pilot navigational problem. Once a plane is off the ground, the plane sees no side wind, if it is flown in "coordinated" flight. The entire air mass around the plane moves (in this case) north at 30 mph. The plane moves through that air mass at 400 mph west.

If you google for info on flying small planes, you'll find that your question is not clear enough to be able to answer accurately. "Flying due west" (ignoring whether is true or magnetic,) can mean flying a heading of due west (in which case the track across the ground is in this case somewhat north of due west)*... or flying a course of due west, in which case, the pilot has corrected for wind and will fly a heading of somewhat south of west to fly his due west course.

But you can make whichever assumptions you want: I suspect that what the teacher had in mind was that the pilot flew a heading of due west, and that the wind was blowing from the south. Therefore, the plane would be blown 30 miles northward for every 400 miles it flies westward. If you draw that up, you will find that you can use Pythagoras's theorem, easily, because west and south are perpendicular.

I think that the expected answer for "the speed of the plane with no wind resistance" is the speed of the plane over the ground. In any case if you do a good drawing and explain your assumptions used to make sense of this, ans then do the math, you should get credit. As practice, walk across the living room, and simulate getting blown off course by a wind from the side, and note the length of your path.

* (again assuming "blowing south" means the wind is from the south -- the standard weather man, sailor, and pilot lingo... rather than blowing southward, which would be the reverse)

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#25
In reply to #15

Re: Algebra 2 Question!!!

11/14/2007 10:39 AM

Blink nailed it.

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#35
In reply to #25

Re: Algebra 2 Question!!!

11/14/2007 3:25 PM

Thanks. It hits on a lot of interests: flying, ambiguous questions, physics, aerodynamics...

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#68
In reply to #15

Re: Algebra 2 Question!!!

04/08/2009 5:18 PM

"If there is no wind resistance a plane cannot fly."

Huh... Really? Then why are airplanes built aerodynamically to reduce wind resistance (i.e. "Drag")? What makes a plane fly is the wings and control surfaces. The form of a wing - or it's "shape". It is designed to create a low pressure over the wing and a high pressure under the wing to create lift.

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#69
In reply to #68

Re: Algebra 2 Question!!!

04/13/2009 2:54 AM

Huh... Really?

Yes, really.

The formulas for lift and drag are essentially the same. Cd (or Cl) x 1/2 x rho x V2 x Area. Rho is the mass density of the fluid (in this case air). The only condition under which there can be no wind resistance (drag) is if Rho is 0 -- meaning the fluid has no mass. In that case, there can be no lift, for the same reason that there is no drag. Without lift an airplane cannot fly.

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#70
In reply to #69

Re: Algebra 2 Question!!!

04/13/2009 8:17 PM

Whatever you say... but I will continue designing aircraft the way it was taught to me. If I use your logic, I'll lose my job.

Checkout: http://en.wikipedia.org/wiki/Lift_(force) - It's explanation is a good primer.

There is another good primer on drag at: http://en.wikipedia.org/wiki/Drag_(physics)

Good Luck.

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#71
In reply to #70

Re: Algebra 2 Question!!!

04/14/2009 10:26 PM

If I use your logic, I'll lose my job.

I assume you are responsible for color selection or perhaps structures.

Every first year aero student knows that there can be no lift without drag. The two articles you referenced (but failed to correctly link) provide the same equations I supplied above. For there to be no drag, the fluid must be massless, or there must be no relative motion between the body in question (a wing, a fuselage, etc) and the fluid. I would hope your exposure to aerodynamics is enough that you can grasp that concept, even if the math baffles you.

If you'd like tangible proof, pick up a copy of Abbott and Doenhoff, and look through the actual wind tunnel data for the numerous wing sections tested. You will find that that there is no wing that creates lift without creating drag. In oversimplified terms, the aerodynamicist's job is to find or develop the wing section and planform that provides the optimum ratio of lift to drag, given the mission of the aircraft -- if you really work around airplanes, you've no doubt heard L/D referenced.

At least you don't sound as daft as CPM, who claims that drag is measured in units of speed: "Therefore, the wind resistance = 1.1234 mph." We can probably agree that such a statement is completely nonsensical.

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#72
In reply to #71

Re: Algebra 2 Question!!!

04/14/2009 10:53 PM

CPM made the reply to your reply. You can always tell when someone is in over their head. They get hostile and insolent. Nonetheless, I (CPM) have no further time for such foolishness and your personal attacks. So here is my final reply to you: Whatever.

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#16

Re: Algebra 2 Question!!!

11/14/2007 3:39 AM

Guest ,

Your teacher has submited clue to you to use pythogorous theorem , just do it and get the answer , if you are a school kid , it is just asignment for pythogorous theorem statement and has nothing to do with wind resistance and air dynamics

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#17

Re: Algebra 2 Question!!!

11/14/2007 4:23 AM

Guest:

You need more information to solve this problem:

1) Is the "400 MPH" ground speed or airspeed?

2) Is "west" the heading or flight path? (We won't get into variation and deviation.)

3) Is the wind really "blowing south" or, more conventionally, FROM the south?

As others have pointed out, the question is meaningless if the airplane has "no wind resistance", but that can't happen anyway. It should also be pointed out that this is a special case in which there is a 90 degree angle between wind and course. Otherwise, the Pythagorean Theorem is not applicable and you'd work with a wind triangle and vectors.

This is a good example of testing in which, if you really understand the subject, you'll fail the test because the person who prepared the test doesn't understand. Your best bet is to try to guess what they were thinking and respond accordingly. Draw the right triangle with sides 400 and 30 and apply the theorem. Good luck.

DickL

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#27
In reply to #17

Re: Algebra 2 Question!!!

11/14/2007 12:54 PM

DickL YOU ASKED THE Q's I wanted to.

Now let us see. @400 W /ward ,thrusted by say To SOUTH@30 would make plane head southward from E>W by a small angle Θ.

tanΘ= 30/400= 0.075~

AutoPilot will reset Rudder to point to final Target at W. Thereby plane will be slowed in the actual E>W Flightpath at actual groundspeed of 400cosΘ=398 mile/hr~~

I AM SLEEPY. The Solar Calculator won't work.But the formula is that.

Schoolkid-please find out the Exact result.

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#66
In reply to #27

Re: Algebra 2 Question!!!

11/29/2007 8:29 AM

No MUKULMAHANT - 400 mph due west is the resultant (ground) speed, not the air speed. To do this speed he needs an air speed √(4002 + 302) = 401.12 mph. Heading is slightly north of west, by atan(30/400) = 4.29°.

This is assuming it's set as a genuine school problem, ignoring the poor wording - see my and various other postings.

Also assumes when he says wind blowing south he means towards the south. Some posters have pointed out convention that wind direction is from, but (I think) you'd call that a southerly wind.

Cheers.....Codey

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#19

Re: Algebra 2 Question!!!

11/14/2007 8:29 AM

I'm really glad that the teacher is a teacher, and not a airline captain! You might look under "crosswind component" in aviation.

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#20

Re: Algebra 2 Question!!!

11/14/2007 8:32 AM

I have read all of the post thus far. Being a physics teacher, these are my comments:

High school kid: I appreciate your tenacity. However, you can find the information you need from just about any website that explains the Pythagorean theorem. Furthermore, that would help us to not just give you the solution. It's difficult to give an explanation in this format that would not just be "giving the solution."

Teacher: You should be able to work this problem if teaching Algebra 2

CR4: It is quite possible that the teacher is not a math teacher and was assigned to fill the position due to a shortage of math teachers. It seems that industry pays considerably better than education. Technically minded people tend to gravitate toward jobs that will pay their bills and feed their families. This leaves a void for qualified math and science teachers. Additionally, most teachers do not have degrees in the field they teach. Instead they have degrees in education.

With all of that said, High school Kid, become a member of CR4 and learn more than you ever imagined possible. Do not let our cynicism discourage you. You have taken on an extra task and should be commended for your effort. However you unintentionally violated the unspoken rules of academia because of your youth (I'm not offended, we have all done this at one time or another). Press on, dig in and work toward the answer and it will stick with you!

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#24
In reply to #20

Re: Algebra 2 Question!!!

11/14/2007 10:39 AM

High school kid: I appreciate your tenacity. However, you can find the information you need from just about any website that explains the Pythagorean theorem. Furthermore, that would help us to not just give you the solution. It's difficult to give an explanation in this format that would not just be "giving the solution."

I think he must know the Pythagorean theorem pretty well -- he could not have gone beyond algebra 1 without it. I suspect where he is lost is in the nature of flight, and the bizarre wording of the question: without wind resistance there can be no flight. Perhaps he is imagining that the wind blows across the airplane as it flies, and that he would need to know what the drag coefficient, projected area, etc. is for a wind from that direction. Even if he were aware that the entire air mass, plane included, moves north or south (depending upon the meaning of the wind direction) he would still be confused by the direction of flight: is "due west" the direction over the ground or the plane's heading?

If he understands the nature of flight, then there are 2 possible answers, (ignoring the little details pilots are concerned with) one which yields a ground speed higher than air speed, and one which yields a ground speed lower than airspeed. But of course, all he should have to do is lay out his assumptions, and do the math accordingly. Unfortunately the "answer" has a 50-50 chance of being based on the assumptions he picked. You, as a physics teacher, could give him full marks based on an understanding of his assumptions. Unfortunately, it doesn't sound like his teacher can understand the question, let alone his assumptions -- so if he comes up with the "wrong" of the two correct answers, then he gets marked wrong -- and the educational system has done him a disservice.

Additionally, most teachers do not have degrees in the field they teach.

Sadly. Actually, just a minor in the subject matter would be a huge help. My daughter had a science teacher who asked, on the first day "Who here doesn't like science?" When about 1/2 the students raised their hands, she said "I'm with you." And she meant it. I dare say that by the end of the year, even fewer of her students would like science.

This situation is mainly the fault of a system we have all created by failing to become adequately involved, I suppose: Clearly, it is nuts that we pay the people who teach our kids a fraction of what we pay a baseball player. But it is also the fault of teachers: if you don't like science, you should have the decency to say, when offered the position: "Sorry, I'm incompetent in that field". I am sure that if someone offers me a job teaching brain surgery, I'll turn it down.

Perhaps a good strategy for the student is to read up on flight planning, wind correction angles, etc in one of the hundreds of books that have been written on preparing for your private pilot's written exam. Then, when he takes his answer to his teacher, he'll have a reference for authority. And, he will have learned a little about flying in the process. Lemonade from lemons.

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#64
In reply to #24

Re: Algebra 2 Question!!!

11/29/2007 8:09 AM

Exactly. It's not enough to just know Pythagoras, need to think about the question and understand the physics of it.

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#21

Re: Algebra 2 Question!!!

11/14/2007 9:05 AM

I have a pretty good feeling your teacher is just wanting u to use pythagorean theorum in the scenario where the plane starts at a point and is pushed north by the south wind. Make a right triangle and fill in your values where u think they fit!!

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#22

Re: Algebra 2 Question!!!

11/14/2007 9:34 AM

I think we are trying to make the question overly difficult, lets remember that this is high school math, not even physics, and that the answer is probably just the output of the theorem. The question is undoubtedly worded poorly, no wonder the teacher can't understand it. Good to see a student branching out and seeking an answer, I know that I probably would have decided I didn't need any extra credit if I were in his/her shoes!

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#23

Re: Algebra 2 Question!!!

11/14/2007 10:28 AM

One the same lines as MrChevy, is the quest getting extra credit for the answer to the question or for getting on the forum with a question?

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#26

Re: Algebra 2 Question!!!

11/14/2007 12:05 PM

Theres a good reason why the teacher can't get it, the question is worded poorly.

The plane IS travelling WEST WHEN the wind is blowing SOUTH. The westward direction is the hypotenuse of the triangle because it is the resultant of the original travel vector and the southern blowing wind vector, which makes this problem impossible because you cannot have a resultant vector orthoganal (that means at 90 degrees, west versus south) to a force acting on it and still have it be the hypoteneuse of the triangle in a standard 2D plane. You can have a triangle with more then 1 right angle when drawn on a 3 dimensional surface (for example, it is possible to draw a triangle on a sphere which has 3 right angles) but that complex of a geometric surface seems well above and beyond the scope of high school algebra. The only way the resultant vector could be travelling at 90 degrees to the direction of the wind is if the force of the wind on the plane is completely negligable, in which case your answer would be 400 MPH.

On a more practical note, you cannot know how the wind resistance affects the plane if you cannot calculate what that resistance is. A 30 MPH cross wind is all fine and good, however 30 MPH against the head of a pin is alot less pressure then 30 MPH against the side of a barn. If it wasn't, sailboats would hoist hankerchiefs instead of giant sails. Without knowing the size and shape of the plane, you cannot determine how mush sideways "push" the plane is experiencing. And since the question explicitly calls into question that wind resistance, again, the question is impossible to solve.

Ignore what these power users and guru pseudo-intellectuals are telling you (or not telling you as the case might be). Apparently they havent mastered the complex action of READING. And shame on them for telling you to "look it up". Obviously you are and that helped lead you here. A place where individuals touting themselves as professionals and intellectuals would seem an ideal place to gain such help instead of wasting time "googling".

Anybody can go to state university.......

Oh, and I'm only a guest beause I forgot my password.

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#29
In reply to #26

Re: Algebra 2 Question!!!

11/14/2007 1:09 PM

Resistance is only important if you're actually resisting it. The plane is swept along in the medium in which it is travelling. That is why there is such an important distinction between wind speed and ground speed. Do you have to know the coefficient of Resistance to know that a boat travels with the current? A boat crossing a river will be swept downstream with the current. If you want to land at a point directly across from your launch, you need to point your boat upstream and pick a velocity that will compensate for what is lost to the current. You will then have a resultant vector that is orthogonal to the bank.

The original question, while poorly worded for the real world, but perfectly reasonable for Algebra 2, has all information needed to derive the answer asked for and even determine the plane's heading, if desired. It is worded for a student's understanding, not an engineer's or aviator's.

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#34
In reply to #29

Re: Algebra 2 Question!!!

11/14/2007 3:17 PM

Give these kids some credit. When I was flying a lot (and I think this is still true) a kid could get a glider license at 14, and a full pilot's license at 16. When I was taking algebra 2 (in the days of kerosene tv) I knew the difference between heading and course, as well as the difference between indicated airspeed, true airspeed, etc.

It think the question would leave many algebra 2 students (but fewer physics students) scratching their heads. If by "flying due west" we mean holding a heading of due west (and we ignore magnetic, true, variation, deviation, etc.) Then the plane flies further than 400 miles in an hour. If by "flying due west" we mean that the pilot has applied the correct crosswind correction to fly a due west course over the ground, then the plane flies less than 400 miles. Certainly, the notion that the plane is carried along in a mass of air (and that that mass creates no sidewind component at the aircraft) is pretty foreign to algebra 2 students -- we even have one respondent here, (in the post to which you are responding) who presumably graduated high school, who does not understand that concept.

In one of my lives I was an instructional designer (in fact, I started that career training airline pilots) and am unusually sensitive to 1. poorly worded and ambiguous questions; but especially 2. teachers who cannot see the ambiguities, and mark the bright kids "wrong". A while ago I manufactured a thread here, supplying many of the "guest" posts, that has to do with this subject of expected answer vs other reasonable answers.

But all that should be required to get an A is to spell out your assumptions. If a controller say "fly 270" it has a specific meaning: fly a magnetic heading of 270. The course over the ground could be 240 -- but the controller knows that, as does the pilot. Ambiguity is the stuff of life.

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#37
In reply to #34

Re: Algebra 2 Question!!!

11/14/2007 3:34 PM

"Give these kids some credit. When I was flying a lot (and I think this is still true) a kid could get a glider license at 14, and a full pilot's license at 16. When I was taking algebra 2 (in the days of kerosene tv) I knew the difference between heading and course, as well as the difference between indicated airspeed, true airspeed, etc."

It is a poor statement of our education system that textbook questions are routinely "dumbed-down" to the understanding of the least-capable instead of worded to educate the inexperienced on the topic at hand. This would have been a grand opportunity to introduce some of these aviation concepts to kids that have never given it a first thought, and thereby increased the fullness of the educational experience. Pity.

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#43
In reply to #37

Re: Algebra 2 Question!!!

11/15/2007 8:17 AM

The system is inches deep and miles wide...

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#44
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Re: Algebra 2 Question!!!

11/15/2007 11:08 AM

Just so.

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#32
In reply to #26

Re: Algebra 2 Question!!!

11/14/2007 2:45 PM

You're certainly not a pilot, and it appears you may have never studied a physics text.

Physics texts are full of questions like this, (albeit less ambiguously worded) using both flying and boats across streams as good introductions to vector sums.

You'll need to spend some time googling to come to grips with the basics of flight. The fact is that the physics are not at all as you have described. Wind resistance "pushing" the plane this way or that simply does not apply. The plane flies within a mass of air which moves relative to the ground, but not relative to the plane. The only wind that blows relative to the plane is "airspeed" which is always perfectly straight over the nose, if the plane is flying and is in coordinated flight.

Here is a computerized replica of the E6B calculator every pilot is familiar with.

BTW PWSlack, from all I have seen, is a real intellectual, not a pseudo intellectual. In other words, he routinely uses his intellect to think things through. We would all do well to emulate his thoughtful approach.

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#28

Re: Algebra 2 Question!!!

11/14/2007 1:05 PM

uh.... just re-read the question. #1 the plane is flying at an IAS of 400 mph now matter which way the wind is blowing. Now, if you're after Groundspeed, that's an entirely different question! It's also going to matter whether or not our 'pilot' is needing to fly due west to his destination, or just boring holes in the sky and not figuring in the cross wind component for his heading. If he keeps the aircraft pointed on a due west heading, in an hour, he'll be 400 miles further out, but 30 miles off course!

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#31
In reply to #28

Re: Algebra 2 Question!!!

11/14/2007 2:29 PM

Actually, the question does not say IAS. That points out yet another ambiguity. To get true airspeed from IAS, you need to know altitude and temperature.

So it is safe to assume that it is true airspeed being referenced. But, as you say, there are two conditions: in one the pilot is sightseeing and holding a magnetic heading of 270. (we can reasonably ignore deviation and variation) In that case he flies the sq rt of 400^2 + 30^2 or 401.1 miles, for a speed over the ground slightly greater than his airspeed.

Then there is the other case, where he is actually trying to get to a particular place, in which case he applies a wind correction angle. From the ground, he would appear to be crabbing into the wind slightly (but from the plane's perspective the apparent wind is always straight onto the nose). Then the wind triangle gets flipped over and his speed over the ground is less than his airspeed.

The first case is rarely calculated by pilots -- you are unwise to simply fly a heading unless you are intimately familiar with the terrain -- it's remarkably easy to get lost. The second case is calculated all the time -- in fact, it is calculated before every trip if the forecast winds are anything but calm.

But a student who is not a pilot could come up with either answer: 401 or 399, and be right either way, as long as he annunciates his assumptions.

Back when I flew a lot, we used an E6B calculator for divining such things. This site has a computerized version which does the same calculations, but without benefit of the E6B's graphical presentation. And of course the E6B, being a specialized slide rule, used no batteries, so you could rely on its working all the time.

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#33
In reply to #31

Re: Algebra 2 Question!!!

11/14/2007 2:47 PM

Ooops, I'd intended to included a link where I wrote "this site" .

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#36

Re: Algebra 2 Question!!!

11/14/2007 3:33 PM

wow -

Agree with Blink, was almost (only almost) tempted to get out my old plastic flight computer from 1970's (ouch dating) Cessna lessons. Lots of effort expended to help here. Wonder if it is worthwhile introducing the concept of 'Spherical Trigonometry' and really screw the student up - since the plane is actually not following one of the straight lines in the p- theorem at all, but rather an arc [if they are maintaining an 'altitude' - most at 400 mph would probably pay attention to that as well] -

I'm not sure that not answering all 'guests' would be any fun at all, perhaps the sign up process could be shortened for the quickies?

Jim

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#41
In reply to #36

Re: Algebra 2 Question!!!

11/14/2007 10:10 PM

OUCH ! ! You hit a nerve, how about 1940's J 3 Cub. For navigating, we followed the old "iron" compass ----railroad tracks (just kidding). E6B, them was the good old days.

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#42
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Re: Algebra 2 Question!!!

11/15/2007 12:30 AM

One of my most memorable flights was in a 1946 J-3 from Pittsburgh PA to Melbourne FL. Part of the time we really did follow railroad tracks, interstates, and the Shenandoah valley. Cold, drafty... sling seats at just the right angle to nearly cause a sex change operation. Accompanied by eagles at times, and often going no faster than trucks on I 95. No lights, no radios other than a hand-held, just a kerosene compass. Two full days of flying dawn to dusk. Real flying.

My FAA examiner was an old corporate pilot with 30,000 some hours in DC3, Beech 18, etc., and he flew loads of old taylor craft, luscombe, stinsons, etc. He had the greatest stories, and could keep a whole room of young pups such as myself riveted, and often in stitches. He was an exception to the rule -- he was old, and he had been, at times, insanely bold.

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#45
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Re: Algebra 2 Question!!!

11/15/2007 1:37 PM

Mountain flying was an adventure.........From Nashville TN to Knoxville TN (3+ hr flight in 50 HP J-3), across the mountains, the old magnetic compass swung like a merry-go-round. Flew a straight line using the 3-point system. Couldn't fly with the eagles, but did do some "buzzard" flying, also a few ducks. Got my ticket in 1940 in Knoxville while at Univ. of Tenn (Civilian Pilot Training). As you say, no radios, no lights; control tower used a light gun to signal, etc....Those were the days when flying really was fun.

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#46
In reply to #45

Re: Algebra 2 Question!!!

11/15/2007 3:29 PM

Wow, 1940. I got mine in 1978, so when I flew the J-3 it was an antique, but looked almost brand new. I haven't flow for almost 20 years (basically since kids) but my daughter wants to learn, so I hope to take it up again in about 2 years, and teach her.

Old F**rt?? You must be just a young pup! If you got your ticket at age 2, you'd only be coming up on 70. Its great to have some wisdom here!

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#47
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Re: Algebra 2 Question!!!

11/15/2007 4:27 PM

Ken, O.F. et al,

I've got to give you guys a big here.

Nice to have you guys reminiscing, but there's a time & place ... .

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#48
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Re: Algebra 2 Question!!!

11/15/2007 5:58 PM

Giulty as charged.

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#50
In reply to #47

Re: Algebra 2 Question!!!

11/16/2007 2:55 AM

ALSO GUILTY AS CHARGED......WON'T DO IT AGAIN.JUST COULDN'T RESIST.........

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#49
In reply to #46

Re: Algebra 2 Question!!!

11/16/2007 2:49 AM

THANKS...........1940 I was a sophomore at Univ. Tn., 19 yrs old (makes me 86). Took time out from college until '46 finished in '48 / BSME. Retired from TX A&M Physical Plant Engineering & Energy Section, living in the country.....deer walking through front yard munching grass and acorns, enjoying great-grandchildren & CR4. Started flying again in '95, Cessina 150 and doing well according to my instructor. Had to quit just before getting my ticket reinstated, eyes plus quadruple by-pass grounded me for good. Life's been a hoot !

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#51
In reply to #49

Re: Algebra 2 Question!!!

11/16/2007 10:34 AM

I'll try to remember to check this as off topic.

Sadly, I find myself at that point in life where I have to advise my father to stop driving. A relative called to say that he'd ridden with my dad, and between the running of stop signs and driving at 3 mph with the people behind honking angrily, he found the experience pretty frightening. (I wish this relative, Bob, had mentioned the concerns to my dad at the time, because Bob is much closer to my dad's age, and it would also fit a bit more smoothly into the conversation, as in "Holy keeeerapp, Dick, you just ran that stop sign!") Oh well... I'll have to dig up my old motorcycle road racing leathers, and a helmet, and go for a ride with my dad.

It seems a shame that you're not flying. You could always fly with an instructor, I suppose, although that adds some cost. Probably flying with a non-instructor pilot would mean handing over the takeoffs and landings to that pilot (I think ordinary pilots, especially friends, would not be likely to take control early enough to avert a problem). I remember being dismayed when Bob Hoover (also 86, now) was denied his medical certificate by the FAA -- he is, after all, a guy who could fly better on death tremors alone that I can fly on a good day.

Bob pouring tea while doing a barrel roll: http://www.alexisparkinn.com/photogallery/Videos/2006-3-11_bob_hoover.avi

Blog re medical certificates: http://blogs.chron.com/lightflight/archives/2006/05/flying_and_well.html

I'd have to say I'd feel safer with you flying somewhere overhead than I would with half of the Atlanta drivers mere feet away.

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#52
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Re: Algebra 2 Question!!!

11/16/2007 5:11 PM

I think the kid is gone now.

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#53

Re: Algebra 2 Question!!!

11/17/2007 8:07 AM

First if the teacher can't calculate the answer to the question what is she doing teaching the subject.

Okay imagine a highway going straight west. You stand there with a helium filled balloon that you bought for your girl friend. Your teasing her when the balloon slips from your fingers. The wind starts to blow it straight south at 30 mph. You know if you don't get the balloon back for her thats the end of it. She's mad hops in your car and takes off driving at 400 mph down the road west. Boy you have a fast car. So here you are running after the balloon and in an hour you finally catch it(your pretty fast too). Now how far where you from your girl friend when you caught the balloon?

The answer in mph is the answer to your question.

bonus question how fast do you have to run to catch her in one hour if she is still heading west. Oh the second you caught the balloon a state trooper pull out behind her. She didn't get stopped but now she doing the speed limit at 75 mph.

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#54
In reply to #53

Re: Algebra 2 Question!!!

11/19/2007 9:18 AM

and the answer is 401.12 MPH. Sorry I'm late, but it's been a busy week.

(I did enjoy all the posts though!)

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#55
In reply to #54

Re: Algebra 2 Question!!!

11/19/2007 11:50 AM

and the answer is 401.12 MPH...

... unless you actually wanted to get to a place due west, in which case you would fly with a wind correction angle, and your speed over the ground would be less than 400, not more. This is actually more typically the case, because 98% of the time you fly to get somewhere, even if you are site-seeing.

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#59
In reply to #55

Re: Algebra 2 Question!!!

11/20/2007 9:27 AM

Back to the old "heading vs. course" question. I agree if the course is due west the final speed is less, if the heading is west the speed is slightly higher.

I heard someplace that 9 times out of 10 wind is the enemy of pilots, due to lost airspeed. It usually is against you.

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#56
In reply to #53

Re: Algebra 2 Question!!!

11/19/2007 5:18 PM

If you have a car and a girl friend what the heck would you be doing chasing the balloon

Del

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#57
In reply to #56

Re: Algebra 2 Question!!!

11/19/2007 5:28 PM

Poor s*d has neither - girl's b*ggered off with the car!

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#58

Re: Algebra 2 Question!!!

11/20/2007 12:12 AM

I'll bet the teacher DOES know how to construct this equation. And the student did not exactly state the question. And I'll bet the student does not understand angular measurement or the pythagoras theorem. Now, let's guide him through the process of construction.

Using compas direction north as 0 deg, and moving clockwise, east is 90, south becomes 180, west is 270, etc. Now, beginning at a point A and heading due west (neglicting deviation, declination), the bearing would be 270 deg. Suspend the plane in a hugh bubble of air and subject it to the 30 mph wind blowing from south to north. At the end of 1 hour, the plane would be 30 miles north of point A. Call this point B. Moving from point B in a direction, the distance representing the (air)speed equal to 400 mph to intersect the original 270 deg line (called the tract) which commenced at point A, and intersecting that line at a point C. You now have a triangle with sides AB -BC - CA. Line AC, bearing 270 deg, is the tract the plane will fly when heading the plane at a bearing represented by the angular measurement from 0 deg (north), clockwise around point B, to the line BC. So, starting at point A, flying AC at heading BC, in 1 hours time you woulf have flown the AC tract. Using the Pythagoras theorum, BC^2 - AB^2 = AC^2. BC = 400; AB = 30, therefore AC = 398.87 miles, actual distance flown in 1 hour (ground speed). We have ignored all the corrections required to produce actual flying data, which is not germane to the problem of identifying the use of the pythagoras theorem.

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#60
In reply to #58

Re: Algebra 2 Question!!!

11/20/2007 1:58 PM

AMEN, BROTHER..........RIGHT ON...........RE: "Wind works against you........." Why do pilots take the airplane off against the wind??? Why do aircraft carriers turn into the wind for take-offs and landings???? One thing you do learn in piloting is winds are not constant and do not always blow in a given direction, which makes "dead reckoning" so much fun..........

g scott

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#61
In reply to #60

Re: Algebra 2 Question!!!

11/20/2007 4:23 PM

"Why do aircraft carriers turn into the wind for take-offs and landings????"

The XO does not like the smell of jet exhausts with his coffee.

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#62

Re: Algebra 2 Question!!!

11/26/2007 2:01 PM

The effect of a perpincicular force acting on an object will not effect the initial speed, only the final trajectory. Therefor the plane is traveling 400mph, however the plane will land off course if the adjustment for the wind direction is not included in the flight plan.

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#63
In reply to #62

Re: Algebra 2 Question!!!

11/26/2007 3:15 PM

The effects are not obvious with high airspeed and low wind speed, but as you can imagine, if a pilot of a small plane flies a heading of due west at 100mph true airspeed, while the wind is 100mph from due north, then the plane will be blown southward by 100 miles as it moves westward 100 miles. Its path over the ground will be 141.4 miles, and its ground speed will be 141.4 mph. Its course over the ground will be 225 degrees (i.e., southwest).

As someone above said, it's the same situation as a canoe paddling at 5 mph across a creek flowing at 5 mph. If the canoer paddles a constant compass heading across the creek's direction, he ends up down stream, and his path across the ground will be sqrt 2 longer than the straight-across distance. (For this to be literally true, you have to ignore wind resistance.)

(This latter problem has practical ramifications for paddlers. Most newish paddlers will aim at the dock across the river, and continually adjust course, so that by the time they are across the river they are paddling almost directly upstream.)

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#67

Re: Algebra 2 Question!!!

04/08/2009 5:10 PM

Pythagorean Theorum: c2=a2+b2 where c2 (C squared) is the hypotenuse or in words: The square of the hypotenuse of a right triangle is equal to the sum of the squares on the other two sides.

Let a = the wind blowing to the south at 30 mph and b = the plane traveling due west at 400 mph.

c2 = a2 + b2

c2 = 30 squared + 400 squared

c2 = 900 + 160,000

c2 = 160,900

c = the square root of 160,900

c = 401.1234

Therefore, the wind resistance = 1.1234 mph. Without wind resistance, the plane's speed is 400 mph unless you want to calculate the wind resistance based on area, the drag coefficient(s), and air density at whatever altitude the plane is flying and what the air pressure or pressure density happens to be. To do so would be idiotic.

Later.

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