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7 Answers

Speed time distance

Asked by: 33437 views , ,
Flight Instructor, General Aviation

An aircraft traveling 100nm at 100kts completes the trip in 1 hour. Now if we add or subtract 20kts to the airspeed why does the time not change the same. For example, at 80kts it would take the aircraft 75min to fly 100nm. Whereas an aircraft traveling at 120kts completes the same 100nm in 50min. My questions is why does the time not remain the same for the 20kt increase or decrease in GS. 

I was asked this question by a chief pilot and he said the 20kt difference was due to an headwind or tailwind off of the 100kts. His answer was it takes a greater amount of time 15min vs 10min because the aircraft is affected for a longer period of time flying at a slower airspeed. 

This does not jive with me. Taking the wind out of the equation you still get the same time 10min for a tailwind and 15 min for a headwind. So in a no wind situation you get the same answers. Therefore, it bust his theory of wind acting on the plane. 

I came to the conclusion the differece is based off the fact that 80kts is 80% of 100kts, where 120kts would be a 83.333333333% difference and that's why the time changes becauuse there is a differnce in the ratio of GS. 

I hope I didn't confuse everyone too much, but I'm trying to understand that concept.

 

 

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7 Answers



  1. John D. Collins on Feb 21, 2013

    Speed equals distance divided by time. Using elementary algebra, time equals distance divided by speed. So the time to go a fixed distance at a speed that is 80% is proportional to the the speed ratio of 100/80 (or 5/4) of the base time. In the same way, at 120% of the speed, the time is proportional to the speed ratio of 100/120 (or 5/6) of the base time. The increase in time for the slower speed is always greater than the reduction in time for the faster speed when using a constant difference in speed. It is for this reason, if you fly a round trip and the winds remain constant, you lose more time with the headwind than you make back with the tailwind.

    A common mathematical error is to attempt to average speeds over a fixed distance. Here is a trick question. You are driving on a two mile course and drive the first mile averaging 30 MPH. How fast must you drive on the second mile in order to average 60 MPH over the entire two mile course. Answer, it isn’t 90 MPH, in fact, it is impossible. To average 60 MPH over a two mile course, the entire course must be driven in exactly 2 minutes. But, it takes 2 minutes to drive the first mile if you are only driving it at 30 MPH. All of your time is used up when you get to the end of the first mile.

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  2. Wes Beard on Feb 21, 2013

    As John stated, it is all about the ratios. At 120 kts you are going 6/5 faster than at 100 kts. It will take (5/6 the time or ~87% ) to travel the same distance.

    At 80 knots, you are going 4/5 faster than at 100 knots. So it will take you (5/4 or 125%) longer than at 100 knots.

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  3. Bob Watson on Feb 22, 2013

    One logical error I saw in the original question, which John touched upon, was where you “take the wind out of the picture.” The problem with that approach is that it changes the question completely. If the question starts as “how does wind affect the ground speed and the travel time” and you take away the wind and your question becomes just “how does ground speed relate to travel time?” In a no-wind condition, all you’re proving is that the 0 wind has 0 effect.

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  4. Jason on Feb 22, 2013

    Bob, Wes, and John,

    Now I’m getting a little confused on what the correct answer is with the whole wind being include or not. Here was the exact question as it was ask to me

    There are two airports which are 100nm apart. If you fly 100 knots how long will it take you to get there?

    Easy it will take 100 hour.

    Great! Now there’s a 20 knot direct wind on your course. Flying TO the destination there is a 20kt headwind and on the return leg there would be a 20kt direct tailwind. How long does it take you?

    I whipped out the trusty E6B and found it to be 1:15 with the head wind and :50 with a tailwind.

    Great! Now why is it that it take 15min longer with the headwind and 10min shorter with the tailwind. Both scenarios are only being increased or decreased by the same 20kts.

    WHY??

    He said because wind effects the aircraft longer while flying shorter. I think it has to do with the ratio in speed percentage difference. However, please disregard our answers and let me know how you would have answered it.

    Thank you again for everyone that is assisting in this

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  5. Brian on Feb 22, 2013

    IMO Wes gave a pretty clear abbreviated version for you Jason. Although the percents are a little off it is purely about the ratios.

    So 100 NM trip in 60 minutes at 100 knots will take 60 minutes. We agree?

    The Math: 100 knots / 100 kts/hr = 1 hr — In this first part the knots on the top and bottom cancel and you’re left with hours for units. The numerical value is 1 because 100/100 is 1.

    1 hr * 60 min/hr = 60 minutes. — In this second step we multiply by 60 to convert our answer to minutes. The hour units on the top and bottom cancel out leaving us with minutes.

    —

    Now we take the same trip of 100 NM and fly it at 80 knots, which takes 75 minutes. Why?

    The Math: 100 knots / 80 kts/hr = 1.25 hours — Like before the knot units cancel and our numerical value of 1.25 comes from dividing 100/80.

    1.25 hr * 60 min/hr = 75 minutes — Again our second step is a conversion to minutes where the hour units cancel.

    —

    Finally the same 100 NM trip at 120 knots. This takes 50 minutes. Again, why?

    The Math: 100 knots / 120 kts/hr = 0.83 hours — As with the others our knots cancel leaving us with hours and the numerical value of 0.83 comes from 100/120.

    0.83 hr * 60 min/hr = 49.8 minutes — As before our second step converts us to minutes where the hour units cancel. We can round this to 50 minutes.

    —

    The difference is in the percentages (the ratios…after all ratios and percentages are one and the same). On the slower flight the percent difference is 25 percent of 60 minutes — or a 15 min difference. On the faster flight, 17 percent of 60 — or a 10.2 minute difference.

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  6. Brian on Feb 22, 2013

    BTW the fraction to percent screw ups here probably didn’t help. You can’t check the math if you have bad numbers. 🙂

    100/120 reduces to 5/6. Well 4/6 is 2/3 is 0.666~ and 6/6 is 1.0. So 5/6 must be equally between 1.0 and 0.666~. Clearly 0.87 isn’t. <3 you guys. Or just use a calculator for this one.

    The other, 100/80 reduces to 5/4. Common, 120%? -_- This one is just simplifying a fraction — 5/4 = 4/4 + 1/4 or 1 and 1/4. Eek

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  7. Bob Watson on Feb 23, 2013

    I liked John’s example of taking it to it’s logical extreme–a wind speed that equals your air speed. If you’re flying at 100mph in a 100mph tail wind, your ground speed is 200mph so you get where you’re going twice as fast. OTOH, if you’re flying into a 100mph head wind, your ground speed is 0, so you don’t go anywhere. If the head wind is even stronger, you go backwards–i.e. you have a negative velocity.

    So, you can’t calculate the time to cover a distance over the ground without working out the ground speed, first (and the ground speed could be negative, in extreme cases). Since most places an airplane starts from or heads towards are on the ground, this accommodation is a natural part of flight planning. When your medium of travel is moving relative to your destination, you need take that into account, which is what all the wind-correction angle and ground speed calculations do. An airspeed of 100 mph matters to the plane and the air. A ground speed of 80 or 120 matters to the pilot who wants to go from one place on the earth to another.

    Thinking in multiple frames of reference takes a little getting used to.

    Walking or in a car, you’re attached to the ground you’re covering so the car and the destinations use the same frame of reference (i.e. the earth). In a plane (and a boat), the frames of reference used by the vehicle are not firmly attached to those of your destination (except, perhaps, if your destination is an in-flight refueling tanker).

    Another example of the same effect, one which might be more tangible, is that of a kayak or rowboat in a river current. Going upstream, you can paddle all day long (if you’re in good shape) and get nowhere, even though you’re going through the water quickly. Your “water speed” is 2-3 knots, but your speed relative to the river bank (the non-moving ground) is 0, or less. Turning around to go with the current, your speed though the water is the same (perhaps less now that you’re tired from paddling against the current), but your speed relative to the river bank is now much faster.

    Now, why does decreasing a speed have a larger influence on the travel time than a similar increase? It’s because the speed is part of a ratio that is used to determine travel time. Travel time is the distance divided by the speed. Because speed is on the bottom (the denominator), decreasing it has a greater influence on the ratio than increasing it does. Consider the difference between 1/4 and 1/2 (reducing the denominator by 2) and 1/4 and 1/6 (increasing it by 2).

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