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The most important part of Archery is the actual flight of the arrow. That is the goal of it all-- to get the arrow into flight.

Initial Arrow Velocity

It is fairly simple to determine the initial arrow velocity. Earlier we determined the Energy stored in the bow to be equal to the work done on it: F x d x e (a factor for the efficiency of the bow)/2. eFd/2 is our stored energy.

We know that the formula for kinetic energy is ½mv^². Now, eFd/2 does NOT equal ½mv^² because part of the bow is moving as well. This amount is represented by k ½ m_bv^². Here, "m_b" is the mass of the moving bow parts, "k" is a small factor equal to the kinetic energies of the bow. The velocity is the same in both the bow parts and the arrow at initiation because of the laws of conservation of momentum.

½ mv^² + k ½ m_bv^² = eFd/2. That's our root equation, and to solve for "v" we get:

v = (eFd/ (m+km_b) )^{^-2} as the velocity of our arrow at its leaving the bow.

Depending upon the angle at which the arrow is launched we get the horizontal and vertical components of the arrow.

V x Cos(Ø) = V_x

V x Sin(Ø) = V_y

At any time during arrow flight those factors can be used to determine the net velocity.

V_y is in a constant struggle with gravity, pulling it downward at a rate of 9.8 m/s^². (Thus it is important to aim above the target). This, however, does not affect V_x in any way at all (and it is V_x which determines the range of the arrow.) Eventually, however, V_y will reach 0 m/s, and decrease after that, falling towards the ground, forming the shape of a parabola.

This is all assume no air resistance, which we wil get into later on.

The distance the arrow travels is equal to Vxt, "t" being the time that the arrow is in flight. To determine the time that the arrow is in flight, we use the equation t = 2(V_y/g). This is a derivative of the acceleration formulation:

v_f = v_i + at

We multiply it by two because. well...what goes up must come down, and while it's coming down it is still moving forward. We multiply this time by the velocity in the x-direction for the horizontal displacement: X = V_xt.

To find the vertical displacement, we simply use Y = (V_yt)/2. "T" here is the total time in flight.

That almost works, if we neglect air resistance--which is not negligible.

We have to consider drag, the fricticious force upon an object moving through a fluid medium.

As you can see, there are two types of drag. SHEAR DRAG is a result of the arrow's moving through the air. Kinetic Energy is expended to move the fluid out of the way so that the arrow can pass. As it pushes on the fluid, so does the fluid push on it (and since it pushes forward, the fluid pushes back--ergo, drag.) Drag is proportional to the velocity of the object and it's size (as you remember.)

As said before, the formulate for drag force is the K-constant of the medium x the velocity of the arrow x the cross-section.

The other form of drag is FORM DRAG, which results in a force pushing perpendicular to the direction of flight when applied along side shear drag. A layer of turbulent, lower-pressure air is caused by the movement of the arrow through the air. This area, the wake, is created using kinetic energy of the arrow itself. This energy creates an apparent force which pushes the arrow upwards; perpendicular to its path of flight.

This force, acting upon the fletching, helps to stabilize the arrow (the fletching are needed because of this force, otherwise the arrow would be force to spin in the air.)

Torque is generated by form drag. Fletchings in the rear of the arrow cause more force to be applied there, resulting in a net counter-clockwise rotation (looking at the picture) which eventually hopes to stabilize the arrow to a horizontal path.

Once this happens, gravity will continue the counter-clockwise torque by drawing the arrowhead downward, and the wake will invert under the arrow, thus the form drag will be pushing the arrow down. This will drive the nose back up, resulting in frequent oscillations in flight (gravity will, however, bring the arrow down again, head-first, once all of the initial vertical velocity has been overcome.)

Larger fletchings will slow the arrow down with more Shear Drag, but will respond better to Form Drag, making the arrow more stable, and increasing accuracy (at the cost of distance.) Smaller fletchings won't slow the arrow down as much, but it is more likely to wobble in the air.