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Tom22

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A helicopter takes 3 hours to fly from Fairview to Ashland and 2 and 1/3 hours to return from Ashland to Fairview. If the wind velocity is 15 knots from the west on both trips what is the airspeed of the helicopter? Ashland is west of Fairview.

Edited by Tom22
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It makes sense as a math problem. You don't even really need to know distance since you know both legs are the same. You can set the rate * time part of the distance formula for both legs equal to each other and solve for rate.

 

2 1/3(k+15)=3(k-15)

7/9(k+15)=k-15

7k/9+105/9=k-15

240/9=2k/9

k=120

 

This doesn't have much applicability as a helicopter pilot problem since you usually know rate and you want to know time, so maybe this wasn't the best forum for it, but claiming it makes no sense or that it's unsolvable isn't a good look when it's just a basic algebra problem. There are lots of opportunities to use algebra and even more advanced math to improve your decision making as a pilot, so I do think it's useful to know this stuff.

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Helicopter pilots tend to be practical. An airspeed indicator is required to be installed and operational before a flight can begin, so a problem requiring solving for airspeed isn't something they tend to care about. Find something with more applicability to actual helicopter flying and you'll probably have more luck.

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Come on guys, obviously the helicopter suffered a bird strike on it's way to Fairview causing the plexi-glass windshield to break, making the aircraft less aerodynamic, increasing the drag and flying at a slower pace than the ASI is showing. The pilot has the screen replaced while he hits on the waitress in the airport restaurant and the helicopter is more aerodynamic on the way back, suffers less drag and makes better time.

 

It's so simple, why over complicate things?

 

-Or-

 

Blame the copilot for screwing up the flight plan.

Edited by zippiesdrainage
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It makes sense as a math problem. You don't even really need to know distance since you know both legs are the same. You can set the rate * time part of the distance formula for both legs equal to each other and solve for rate.

 

2 1/3(k+15)=3(k-15)

7/9(k+15)=k-15

7k/9+105/9=k-15

240/9=2k/9

k=120

 

This doesn't have much applicability as a helicopter pilot problem since you usually know rate and you want to know time, so maybe this wasn't the best forum for it, but claiming it makes no sense or that it's unsolvable isn't a good look when it's just a basic algebra problem. There are lots of opportunities to use algebra and even more advanced math to improve your decision making as a pilot, so I do think it's useful to know this stuff.

You can do math???

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C'mon folks...it's about HELICOPTERS and some other things that were posed to intentionally throw you off.

This really is not a nav or math problem, IT'S A MONEY PROBLEM; therefore, a solution is unlikely.

 

 

-WATCH FOR THE PATTERNS, WATCH FOR THE WIRES-

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