Here is a wind speed puzzle. Your course is due north, 000 degrees. That is you will proceed in a straight line over the ground to a point that is due north of where you are now. The wind is blowing from 100 degrees, that is from mostly off your right wing, but slightly behind you. With no wind at all, your groundspeed would be 100 knots. Will your groundspeed in the described conditions be greater or less than 100 kts?
My answer and why I mention it, in a couple of days. And as usual, if you know you know, give the student pilots a chance to answer first.
Also, I stood on the wing to fuel tonight, then as I slid off the leading edge holding the fuel nozzle I managed to bash my leg on a nacelle fairing. A customer watched me do it, and I was telling him the truth when I smiled and said I was fine, but I have a really amazing colourful bruise on my leg now.
So what's the wind speed? I think we need to know that...
Feel free to answer as a function of windspeed if you need to. Note that I didn't ask for a numerical answer, just "greater" or "less."
Technically it will depend on the strength of the wind. If its a weak one, it'll give you a slight tailwind I think. However, if its stronger, say more than 25 kts, i think it will actually decrease your groundspeed in the given direction because your crab angle to maintain course will start affecting your speed since you are no longer pointed away from thewind
Better rest your leg, elevate it, and apply ice two or three times a day if you can. Have a look at www.mayoclinic.com/health/first-aid-bruise for more details.
The speed answer is "less"
Regardless of what the wind speed is, since it's blowing from the right and slightly from the back, my answer would be your ground speed would be GREATER thank 100kts. It's like it's pushing you forward, isn't it?
-keisha4, fresh PPL
Oh, and goodluck on the bruise. =)
the url I gave you needs /FA00039 tacked on to the end. It should be:
Please forgive me for being too lazy to do the trigonometry properly.
Anyway, since the wind is from 100, and my course is 000, the crosswind component of the wind is significantly greater than the tailwind component.
That means that I have to "spend" more airspeed correcting for it than I gain from the tailwind.
Therefore, my ground speed is less than it would be in still air.
It's a nice game, this quiz.
Let's do another one sometime.
Just to play the game. I only read the first comment and Aviatrix' answer to that then stopped and got out the E6B. I think Geoff is right... we need the wind speed. By my (admittedly crude) calculations after about 30 or 40 knots wind speed the answer changes from "higher" to "lower"
err also, I was thrown by the use of 000 over 360 so simply used 360. Was that incorrect? I've never really heard 000 used so assumed it's a regional convention.
I am not even a student pilot yet, but I'll give a start of the solution:
I'll use v denoting the windspeed
The tailwind component of the wind is -v * cos 100 (deg)
You'll lose useful speed to the tune of 100-sqrt(100*2-y^2) because of crosswind correction (with y=v*sin 100 (deg))
I'll now give you 24 hours to post a conclusion.
I admit that as a sailor I suggested to drop the anchor once to make most progress. ;-)
You highlighted course - a magnetic value whereas wind direction is true, so wind from 100 degrees with a course of 360 would give a slight nose component from the right.
Or, in words of one syllable, less.
Oh, oh, I know....
Your course is 100. Immediately upon contacting ATC, they vector you to 060 for faster traffic whereby you ground speed drops by -v(cos30), or slower.
As you continue, ATC askes you to descend for the airline traffic approaching the busy class B above you. The winds are different at this altitude--directly on your nose or -v(cos 0), or much slower.
"We'll have you back on course in a couple of miles," says ATC.
That controller's time on station has reached it's limit and the next time you hear the voice on the radio, it's a new controller for that sector.
Time goes by and it's apparent the new controller wasn't briefed on your request. You ask for higher to get back to the more favorable winds and the controller says, "standby."
The next thing you hear is, "...I have a new routing for you, advise when able to copy." Being the attentive pilot, you immediately say, "ready to copy."
Your new routing is mostly into the winds for the remainder of the trip at the bumpiest altitude open for this direction of flight.
How did I do?
Oh, I tried all day to not answer this.
If we assume Ws < As ( windspeed < airspeed ), which seems reasonable...
Without an E6-B, the trickiest parts is the sign and direction convention. Wind relative to the course is the "windtrackangle", -280 ( 180 degrees from the 'wind from' convention, sign flipped by vector convention wta = crs - wind )
Then the wind correction angle and ground speed are easy to see
Wca = arcsin( ws/as * sin(wta) )
gs = as*cos(wca) + ws*cos(wta)
So windspeed, windcorrection angle in degrees and groundspeed are:
ws wca gs
0. 0.00 100.00
5. 2.82 100.75
10. 5.65 101.25
15. 8.49 101.51
20. 11.36 101.51
25. 14.25 101.26
30. 17.18 100.75
35. 20.16 99.95
40. 23.20 98.86
45. 26.31 97.46
50. 29.50 95.72
55. 32.80 93.61
60. 36.22 91.09
65. 39.80 88.11
70. 43.58 84.60
75. 47.61 80.44
80. 51.98 75.48
In the language of Backus:
drad = 3.14159265/180.0
wta = -280
as = 100
write(0,fmt='(3a6)') "ws", "wca", "gs"
do i = 0, 80, 5
ws = float(i)
wca = asin( ws/as * sin(wta*drad) ) / drad
gs = as * cos(wca*drad) + ws * cos(wta*drad)
write(0,fmt='(f6.0,2f7.2 )'), ws,wca,gs
I came up with the same results as Jinksto and Sarah: The wind helps slightly, up to a wind of about 34% of airspeed, above that, it is a hindrance.
As A^2 said, there is little to add to Sarah's table. At low wind speeds a small tailwind component, at higher speeds the crosswind correction kicks in.
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