Archery - HS Physics

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Critically|Ashamed
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Archery - HS Physics

Postby Critically|Ashamed » Sun Sep 11, 2011 12:19 am UTC

Hey guys; I need some clarification. I'm doing an experiment based on archery and the range/flight time of the arrow in relation to angle of projection. From what I understand:

The force the bow applies to the arrow acts in horizontal and vertical components; Ux = Vcos(a) and Uy = Vsin(a) where a is the angle of projection, V is the velocity the arrow produces at the given draw length. This, however, would suggest maximum range is at a= 90*. Obviously that's not correct. From experimental data the max range is at a= 45 which suggests Ux = Vcos(2a) --> Vx and etc. I've tried googling it but all I get is cheap archery products and discussions on other facets of archery. Can anyone clarify what the equation should be?

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cpt
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Re: Archery - HS Physics

Postby cpt » Sun Sep 11, 2011 12:42 am UTC

Critically|Ashamed wrote:Hey guys; I need some clarification. I'm doing an experiment based on archery and the range/flight time of the arrow in relation to angle of projection. From what I understand:

The force the bow applies to the arrow acts in horizontal and vertical components; Ux = Vcos(a) and Uy = Vsin(a) where a is the angle of projection, V is the velocity the arrow produces at the given draw length. This, however, would suggest maximum range is at a= 90*. Obviously that's not correct. From experimental data the max range is at a= 45 which suggests Ux = Vcos(2a) --> Vx and etc. I've tried googling it but all I get is cheap archery products and discussions on other facets of archery. Can anyone clarify what the equation should be?


It might help your googling to search for the concepts of projectile motion in general, and not specify archery.

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thoughtfully
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Re: Archery - HS Physics

Postby thoughtfully » Sun Sep 11, 2011 12:47 am UTC

Leave archery out of it and search for projectile motion or trajectories adding "maximum range" or such will probably prove useful. Hyperphysics is a great resource for this sort of thing.

Remember, you are maximizing distance (specifically, horizontal distance; vertical distance doesn't count!), which is velocity times time. When you raise the elevation of the arrow with respect to the horizon, you gain some extra flight time, and that gives you some more horizontal distance, before it reaches the end of its flight. But if you elevate it too much, it spends most of the time in vertical motion, which doesn't add to the range.

So, for maximum range, you need to balance the horizontal velocity with the time of flight. This seems to be where you are confused.
Last edited by thoughtfully on Sun Sep 11, 2011 1:09 am UTC, edited 1 time in total.
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iop
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Re: Archery - HS Physics

Postby iop » Sun Sep 11, 2011 12:56 am UTC

The time of flight depends on how much time the arrow spends in the air. An angle of 90 degrees maximises this. The distance the arrow flies is the product of horizontal speed and time of flight. At an angle of 90 degrees, the horizontal speed is zero, so the range is zero. Thus, you want to calculate the time of flight as a function of angle, and then multiply that with horizontal speed, which is also a function of the angle. That will allow you to find the optimum.

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Dopefish
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Re: Archery - HS Physics

Postby Dopefish » Sun Sep 11, 2011 12:57 am UTC

If Ux and Uy are meant to be the horizontal and vertical components of velocity, then Uy will want a -gt term to account for gravity. If on the other hand U is meant to be the final position of the arrow, you'll notice they don't have the right units, currently having units of velocity.

Also, you'll probably want to account for the ground, which will presumably stop the arrow as soon as it hits it, and as such t_max would be when the arrow cross the x axis assuming level ground and initially firing from the origin. (You could twerk things to account for the arrow being fired from a point above the origin, or unlevel terrain will a little bit of extra effort too.)

Realisticly, the optimal angle isn't actually apt to be 45 degrees either, due to air resistance, but neglecting that then 45 degrees is correct, but you should be able to derive that.

The explicit equations for the position/velocity and such are derivable at the high school level and so I'm hesistant to state them outright, but if you are looking to google things I wouldn't specificly google for archery information, but general projectile motion information, since those equations are fairly general.

Critically|Ashamed
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Re: Archery - HS Physics

Postby Critically|Ashamed » Sun Sep 11, 2011 1:29 am UTC

Uy is the initial Y-axis velocity (that is, straight up). The final displacement of the arrow in the vertical axis is 1.73 (my height). Most of that is undercontrol, it's more proving why the maximum is at 45. I think using the Horizontal speed * time method will be the most beneficial for this.

I've been using standard newtonian motion formulas for what you mentioned Dopefish; Sy = ut + 1/2 at^2 (a being gravity). I think Iop's method may be the best way to handle this; I'll try that and see. Thanks guys.

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Tass
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Re: Archery - HS Physics

Postby Tass » Mon Sep 12, 2011 9:24 am UTC

Critically|Ashamed wrote:Uy is the initial Y-axis velocity (that is, straight up). The final displacement of the arrow in the vertical axis is 1.73 (my height). Most of that is undercontrol, it's more proving why the maximum is at 45. I think using the Horizontal speed * time method will be the most beneficial for this.


If you include you height the optimal angle will actually not be exactly 45 degrees, but since the arrow is able to fly much higher than your 1.73m it wouldn't make much of a difference, to little for you to detect it experimentally.

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idobox
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Re: Archery - HS Physics

Postby idobox » Mon Sep 12, 2011 12:43 pm UTC

The procedure is quite simple.

Express the initial velocity for x and y, then, using the vertical component, find the flight time before the arrow touches the ground as a function of the angle, then use this time with the horizontal component to find the distance travelled.
Then, you have to find the maximum of this function, by deriving.

If you suppose initial height is 0, and neglect air resistance and other effects, the best angle is 45°
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Roger K
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Re: Archery - HS Physics

Postby Roger K » Sun Sep 25, 2011 5:44 pm UTC

By the way, if you can disregard air resistance, say on an airless planet, the trajectory would be nearly a parabola, actually a segment of an ellipse, but on a sufficiently large planet, the difference is moot. The arrow would tumble because air resistance on the fletching (tail feathers) is what keeps it straight. In the normal case, you can't disregard air resistance, which depends on air density, the coefficient of drag (which is different for different shaped arrows), and velocity squared. Air density and coefficient of drag can be treated as constants which may be determined by experiment, but since the decelleration changes as the arrow slows down, the formula for the flight path involves analysis that I am not prepared to do at this time (actually not at all).

I have seen a contest amongst archers for the longest distance. The winner could not say what angle she shot the arrow, but could tell me the relation between the lower limb of her bow and the horizon. She had determined by experiment the relationship to maximize flight distance.


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