Monday, December 3, 2012

Determining a Close Approximation of Parabolic Arc Length

...without all the mucky-muck of confusing calculus. maybe one day i will learn to comprehend and love the above formula for finding arc length, but for now i don't understand it (and i have tried to, oh lord, how i have tried) and i cannot bear to look at it. that integral symbol belongs on a violin...not in a math equation!

if only you could actually plug in some readily available numbers for dy and dx and have an answer pop out the other end....but noooooo, you have to have some prior esoteric knowledge of something else.

here's how i feel about it...

and now...on with an intuitively easier way

thank you to RolandMoutal on GeoGebra for creating a program which has allowed me to make this illustration. here is the actual interactive graphic

in the above example:

if an object has a velocity of 90 m/s at an angle of 37° then the velocity for x and y will be:

Vx = 90 cos 37 = 71.877 m/s
Vy = 90 sin 37 = 54.163 m/s

The maximum height of the object can be determined by using the following steps:


Velocity =  gt
where g = gravity 9.8m/s2 and t = time

plugging in the numbers we find:

54.163/9.8 = 5.527 seconds
the time it will achieve maximum height

then we use this formula to determine the position:

position = Voyt – ½gt2

54.163 (5.527) – 4.9 (30.546) = 149.68 m

what goes up, must come down at the same velocity and time it took to go up.
So it will take 5.527 seconds x 2 to go up and down = 11.054 seconds

to determine x-distance we multiply the x-velocity (71.877 m/s) times 11.054 seconds = 794.512 m

arc length:

for a close approximation of arc length, divide the x-distance (i.e., the range) and the time by a minimum of 20.
794.512 m/20 = 39.726 m
11.054 seconds/20 = .553 seconds

next determine the y-postion at those time intervals.
click for larger animated gif image
for example:

at .553 seconds (position 1 on the diagram below) the position of y will be 28.439 m
at 1.106 seconds (2 on the diagram) the position of y will be 53.884 m
at 1.659 seconds (3 on the diagram) the position of y will be 76.335 m
at 2.212 seconds (4 on the diagram) the position of y will be 95.793 m
and so on up to 5.527 seconds.

(you only need to do calculations for the first half of the arc, because the numbers for the second half are exactly the same...or as peter noone from herman's hermits might say, "second verse, same as the first! i'm a parabolic arc i am, parabolic arc i am, i am..." well, you know the rest)

The position of y will be determined by using the above formula for position.

Then we must find the length of each of those y-segments by subtracting the y-position below from the y-position above.
For example:

95.793 – 76.335 = 19.458 ( position 4 - 3 on the diagram)
76.335 – 53.884 = 22.451 (3 - 2)
53.884 – 28.439 = 25.445 (2 -1)
and the first length of y is 28.439 (position 1 on the diagram)

the x-axis segments will be constant at 39.726 m.

once you have determined all the y-lengths, you can then calculate the hypotenuse for each of the 10 little triangle segments of the first half of the arc by using the pythagorean theorem:

a2 + b2 = c2
or in our case:
x2 + y2 = hypotenuse

calculations for arc length when the range is divided by 20
for example:

√39.7262 + 28.4392 = 48.856m
√39.7262 + 25.4452 = 47.176m
√39.7262 + 22.4512 = 45.631m

and so on for all ten.

Once you have completed that, add up all ten and then multiply by 2 to achieve the complete length of the arc.

In our example, after all our calculations we should achieve an arc length of 864.182 m.
this is very close to the actual arc length of 864.327 we find by using the segment of a parabola calculator.

*note – to get even closer to the actual amount you can divide the x-distance and time into more segments.
Try 40, 50 or more (see below). But for all practical purposes, 20 works fine and will get you to within a few centimeters of the true arc length.
calculations for arc length when the range is divided by 40...getting warmer!

incidentally and fyi, the formula to find the area under the parabola is 2/3wh, i.e., width times height times .666. this is also affectionately known as the parabolic formula of the beast.


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