A Little Help With an E6-B Wind Problem?

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turbojet28

Well-Known Member
(This post may belong in Technical Talk, but I figured I might get more of a response here)

I'm currently doing a little manual E6-B quiz/exercise to help me prepare for trying out for WMU's NIFA flight team this coming fall, and I have run into a slight problem. The question is this:

True Heading: 253º
TAS: 140 KTAS
Wind: 123º at 21kts.
Find true course and groundspeed

Is this similar to figuring winds aloft using GS, TC, TH, and TAS? I can do those just fine, but I can't seem to figure out how to do this one. Anyone with any advice? I can do it using vector sums and trigonometry, but can't figure it out on the manual E6-B. Thanks!
 
Set Grommet at 100kts

Make sure the Wind direction is under the true index

Mark the grommet at 100, Mark a spot above the grommet that reps the wind speed e.g. 25kts wind speed, mark 125kts

Let's call the spot marked over the grommet, the wind dot.

Put the true heading under the true index

Slide the wind dot to the true airspeed if you have that number and the grommet will point at the groundspeed

Use the wind dot and note how many degrees that it is off center, it is off to the left, subtract the degrees from the heading and if it is on the left, add it.

Use your map to determine magnetic variation and if you have a steer card, use that to correct for the specific compass in use.

I really hope this helps. If it does not, feel free to send me a private message.
 
If true heading is given, then it's essentially a backwards wind triangle. Normally you have true course and need to find true heading with the proper wind correction angle.

And this DOES belong in Technical Talk, and you should give yourself twenty lashings with a wet noodle for trying to beat the system for your own gain!
 
A wet noodle, eh? Will a towel work?!
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Future: Thanks. That helped a lot. The more I play around with an E6-B, the more I realize it can do much more than ever knew. Throughout my PPL, I pretty much just did the basics of press alt., density alt. TAS, conversions, and winds. I guess I just never really paid any attention to what else you can really do with it.

Now that were here, does anyone know how to use it for radius of action problems?
 
What were the available answers, if any, for the last question?

Radius of action: First you need to groundspeed for both the inbound and outbound leg, label them gs1 (outbound) and gs2 (inbound). Next you need to know how much time is available. On these tests, they like to give you the amount of fuel available and a burn rate and will normally state whether or not to include a reserve. Line up (gs1+gs2) over total time available. After they are lined up locate gs2 on the outside, or "a" scale. Under the gs2 value will be time to turn. A lot of the time the questions just want time to turn. However, if the question is asking "how far can you go before you must turn around" then put gs1 under the rate arrow and go to the "time to turn" on the "b" scale . I hope this helps some, if the way I explained it is too confusing just let me know and I will try to clarify. Good luck

(gs1+gs2) / total time (fuel).......go to.......gs2 / time to turn
 
[ QUOTE ]
If true heading is given, then it's essentially a backwards wind triangle. Normally you have true course and need to find true heading with the proper wind correction angle.

And this DOES belong in Technical Talk, and you should give yourself twenty lashings with a wet noodle for trying to beat the system for your own gain!

[/ QUOTE ]
That's how I would think to do it, however if you try it on the E6-B it dosn't work. I set the wind mark, 123 at 21, then rotated 253 under the true index and the wind mark on 140 and it is 7 degrees left and I get a groundspeed of about 152. Since we already have true heading I would add the 7 degrees instead of subtract to get true course and come up with 260. Sounds about right but if I test the answer by setting the true course of 260 under the true index and keep the wind dot on 140 I goes to 6 degrees left and groundspeed moves up two notches making my true heading 254 (260 minus 6 degrees left) and groundspeed 154. Its really close but I don't think it works becaue no matter what combination of true course and groundspeed you come up with you will always have to move it to cross check the answer and that changes true heading and groundspeed. In other words, it always works one way but not the other and if it did work this way it should always come out the same both ways. It also seems that there would be more than one combination of true course and groundspeed that would yeild the same true heading and TAS for a given wind. It almost seems like a trick question becasue you need either groundspeed or true course. If I was given groundspeed I could come up with true airspeed and true course, or if I was given true course I could come up with true heading and groundspeed, but it seems there's not enough information to get true course and groundspeed. Unless I'm missing something.
 
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[ QUOTE ]
If true heading is given, then it's essentially a backwards wind triangle. Normally you have true course and need to find true heading with the proper wind correction angle.

And this DOES belong in Technical Talk, and you should give yourself twenty lashings with a wet noodle for trying to beat the system for your own gain!

[/ QUOTE ]
That's how I would think to do it, however if you try it on the E6-B it dosn't work. I set the wind mark, 123 at 21, then rotated 253 under the true index and the wind mark on 140 and it is 7 degrees left and I get a groundspeed of about 152. Since we already have true heading I would add the 7 degrees instead of subtract to get true course and come up with 260. Sounds about right but if I test the answer by setting the true course of 260 under the true index and keep the wind dot on 140 I goes to 6 degrees left and groundspeed moves up two notches making my true heading 254 (260 minus 6 degrees left) and groundspeed 154. Its really close but I don't think it works becaue no matter what combination of true course and groundspeed you come up with you will always have to move it to cross check the answer and that changes true heading and groundspeed. In other words, it always works one way but not the other and if it did work this way it should always come out the same both ways. It also seems that there would be more than one combination of true course and groundspeed that would yeild the same true heading and TAS for a given wind. It almost seems like a trick question becasue you need either groundspeed or true course. If I was given groundspeed I could come up with true airspeed and true course, or if I was given true course I could come up with true heading and groundspeed, but it seems there's not enough information to get true course and groundspeed. Unless I'm missing something.

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my thoughts also...hence "what are the available answers" and "is that all the information given"
 
In thinking about it, isn't what the question is actually after is groundspeed and track, not true course? That makes more sense as a question, given the data provided.

And that's the first time I've ever heard of "radius of action", much less a method of computation. Is that a commercial or ATP written sort of thing?
 
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In thinking about it, isn't what the question is actually after is groundspeed and track, not true course? That makes more sense as a question, given the data provided.

And that's the first time I've ever heard of "radius of action", much less a method of computation. Is that a commercial or ATP written sort of thing?

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Radius of action is the distance you can fly out, return, and land with reserve fuel remaining.

Point of No Return is similiar, but is point where you can reach, return, land, and flame out on the runway, essentially. Beyond this point, you can't make it back.

Equal Time Point is the point to which, based on winds mostly, that you can fly out to, and either return to departure point, or continue to destination. Not normally a half-way point, but as mentioned before, varies on atmospheric conditions.
 
flyguy, I actually am having the same problem as you now. Earlier when I first read Future's post I only had a couple seconds, quick whipped out an answer and got the same as you, which is close to one of the possible answers. I just figured the errors came in my haste of doing the problem, but now I see where you are coming from as I have gotten the same thing. The choices are (true course and groundspeed):

a) 259º at 154 knots
b) 247º at 162 knots
c) 299º at 137 knots
d) 231º at 166 knots

I chose answer A as the closest to mine, and that is supposedly the correct answer according to the answer key. What I have given above is all the info given. I don't see this as a very practical problem for flying, as true course and groundspeed are what you should know in the first place.
 
Jay, Thanks for the explaination of radius of action. I just have one problem, however. There is no given "total time". The actual question is this:

True Course: 100º
TAS: 95
wind: 100º at 12
useable fuel: 50 gal w/ a burn rate of 9 gph
How long can you fly outbound before you must turn back in order to arrive at your starting point with no fuel left?

The GS outbound is obviously 83 and out is 107 as the wind is directly aligned with the course, but I'm stuck there.
 
Okay, I just broke out the whiz wheels (both E6B and CR-3) and came up with the same answer on both: true course of 260, groundspeed of 154. I'm puzzled as to why you guys are re-calculating groundspeed off the true course; groundspeed is a product of heading, airspeed, and wind direction & velocity. True course doesn't factor into it. Am I missing something here?
 
[ QUOTE ]

True Course: 100º
TAS: 95
wind: 100º at 12
useable fuel: 50 gal w/ a burn rate of 9 gph
How long can you fly outbound before you must turn back in order to arrive at your starting point with no fuel left?

[/ QUOTE ]

You can find your total time by the normal time fuel and distance method...rate arrow on the 9, find 50 on the outside, time will be right below...total time equals about 332 minutes. So as you mentioned the ground speeds are 83 and 107, a sum of 190. Put 190 over (on the outside) 332 minutes (inner scale). Keep them lined up and on the outside scale find 107. Right below the 107 should be 188. Therefore you can fly outbound for 188 minutes or 3 hours 8 minutes (approx.) Again if this is unclear I can try to explain it a better way..just let me know..good luck

x/y x being outer scale y being inner
190/332....go to.....107/time to turn (answer)
 
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Okay, I just broke out the whiz wheels (both E6B and CR-3) and came up with the same answer on both: true course of 260, groundspeed of 154. I'm puzzled as to why you guys are re-calculating groundspeed off the true course; groundspeed is a product of heading, airspeed, and wind direction & velocity. True course doesn't factor into it. Am I missing something here?

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What I was doing was checking my answer. Since I came up with a true course of 260, I figured I'd plug that into the E6-B, calculate for true heading and see if the answer is 153 which it wasn't. Also, once you rotate the dial to your calculated true course you have to slide the wind mark back to 140 because it moved. No matter how you slice it groundspeed is going to change with this type of problem if you re-calculate. In theory if you were to recaluclate to check your answer the groundspeed should not move and you should calculate the true heading to be what it was in the given problem. But it dosn't seem to work that way.
 
Aloft,

Just re-read your post and I understand what you mean now about groundspeed. Yeah that makes sense, but it still dosn't change the fact that with a true course of 260 your wind mark is only 6 degrees left making your true heading 154 not 153. Again its close but it seems to me that if you could calculate it this way it should be the same, but no mattter what you do that wind mark is going to move when you put that the true course under the true index and that is going to change the true heading.
 
Just a guess, but I think that the reason it doesn't work "backwards" is that the crosswind component is greater for 253 than for 260.
 
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Just a guess, but I think that the reason it doesn't work "backwards" is that the crosswind component is greater for 253 than for 260.

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Ahhh, true true. Good point. So I guess htat means the true course is not really quite 260 then, which makes sense as the best available answer (answer A) is 259 and 254. We already determined that 254 is a correct groundspeed so I guess its not a bogus question after all. It is a bogus scenerio though if you ask me.
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Maybe the best way to solve this type of question in the future would be to plug each anwer into the E6-B and see which one works, becuase I just did that for answer A and it works perfect. Would have saved myself a couple days had I just done that to begin with.
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