Riddle me this Batman II

[Super 80]

On a day with absolutely no wind, a pilot flies from point A to point B in a Tomahawk, lands, has a cup of coffee, reads the paper, takes a dump, smoozes with the girl behind the desk, checks out the jet in the hangar and flies back to point A carrying bank checks.

On the next day the pilot repeats the same route at the same speeds except this time he encounters a 50 kt headwind on the first leg and a corresponding 50 kt tailwind on the return leg.

Assume all flights are straight line, the wind is constant and ground times are the same.

Is his total flight time on the second day:

a. less than the first day
b. the same as the first day
c. more than the first day
d. not enough information to determine

 

[snoopy_1]

C is my final answer.
The tailwind on the second leg won't make up for the headwind on the first leg, whenever there is wind it will take longer than if there is no wind.

 

[Quillpig]

C it is indeed. Because you go faster with a tailwind, you spend less time in it, so the headwind has more time to act on you and hence more effect.

Of course, with a 50 knot headwind in a Tomahawk, would you ever return?

QP

 

[TonyC]

C.

I used to know how to "show the work"... I'll have to ask my High School Junior daughter to help

The penalty of the headwind is applied for a longer time than the benefit of the tailwind is enjoyed over the same distance, hence the total time is greater.

 

[VNugget]

1 = a no-wind leg
3 and 4 = headwind and tailwind legs respectively
s=groundspeed
t=time
d=distance
w=wind speed


s = d / t

t1 = d / s

t3 = d / (s-w)
t4 = d / (s+w)

Time for first trip = t1*2
t1*2 = 2d/s

Time for second trip = t3+t4
t3+t4 = [d/(s-w)]+[d/(s+w)]
t3+t4 = [d(s+w) + d(s-w)] / (s-w)(s+w)
t3+t4 = d(s+w+s-w) / s^2-ws+ws-w^2
t3+t4 = 2sd/(s^2-w^2)

If you plug in 0 for w, you can cancel out an s from the top and bottom from the equation for second trip, and are left with the same equatoin as that for the first one.

If you have any w of greater than 0, you can see the denominator shrinking, which means the value of the equation is growing bigger than the no-wind one.

 

[fLYbUDDY]

I don't know but if he has to take a dump again, the second day, I hope he has baggie or somethin!

 

[GogglesPisano]

C

And I finally figured out the balloon one too!

 

[TonyC]

If you have any w of greater than 0, you can see the denominator shrinking, which means the value of the equation is growing bigger than the no-wind one.

I was coming back to post my proof, and found that VNugget beat me to the punch. The numbers I crunched look very similar.

Just to add a fine point... If you have any number "w" (in VNugget's proof used to represent wind speed) OTHER than 0 (zero), that is greater than OR less than 0, the denominator shrinks. This is due to the fact that w is squared.

This means the w in the equation can represent either a headwind or a tailwind, and the problem might have the headwind in the first leg and tailwind second, or the reverse. Both scenarios increase the total time.

 

[Super 80]

Tony,

That's right. VNugget has gone way past my ability to read his math, but his conclusion is one I can understand.

Any wind input in a recipocal out and back arrangement will increase the total time.

If you don't understand the math, then think about it this way: the headwind will affect you for a longer period than the tailwind.

So overall, any wind component will increase the out and back time required to complete an A to B to A sequence.

This also works for a car driving a constant 60 mph and a truck that can do 50 uphill and 70 downhill where the hills are symmetrical.

 

[Super 80]

1 = a no-wind leg
3 and 4 = headwind and tailwind legs respectively
s=groundspeed
t=time
d=distance
w=wind speed


s = d / t

t1 = d / s

t3 = d / (s-w)
t4 = d / (s+w)

Time for first trip = t1*2
t1*2 = 2d/s

Time for second trip = t3+t4
t3+t4 = [d/(s-w)]+[d/(s+w)]
t3+t4 = [d(s+w) + d(s-w)] / (s-w)(s+w)
t3+t4 = d(s+w+s-w) / s^2-ws+ws-w^2
t3+t4 = 2sd/(s^2-w^2)

Okay, now I get it, it took some time to work throught the math, but it's a really nice equation and it does work out well. One thing that threw me was your assignment of s as groundspeed. I think s should be just speed and t1 = d / s + 0.

This doesn't change any of your calculations now that I understand your ^ symbol to be the raised square symbol.

Overall, very well done; I like it, and your proof shows any absolute value of w greater than zero increases t total.

 

[VNugget]

Yup, it’s actually a lot easier if you do it on paper, then you can see the fractions clearly and not get into “parenthetical hell.” (Ever tried typing the density altitude equation into a calculator? Oh boy…)

And thanks to TonyC for pointing out that the square would force w to be positive in any case… doh!