Aerodynamic properties of an archery arrow

Abstract

Two support-interference-free measurements of aerodynamic forces exerted on an archery arrow (A/C/E; Easton Technical Products) are described. The first measurement is conducted in a wind tunnel with JAXA’s 60 cm Magnetic Suspension and Balance System, in which an arrow is suspended and balanced by magnetic force against gravity. The maximum wind velocity is 45 m/s, which is less than a typical velocity of an arrow (about 60 m/s) shot by an archer. The boundary layer of the arrow remains laminar in the measured Re number range (4.0 × 10³ < Re < 1.5 × 10⁴), and the drag coefficient is about 1.5 for Re > 1.0 × 10⁴. The second measurement is performed by a free flight experiment. Using two high-speed video cameras, we record the trajectory of an archery arrow and analyze its velocity decay rate, from which the drag coefficient is determined. In order to investigate Re number dependence of the drag coefficient in a wider range (9.0 × 10³ < Re < 2.4 × 10⁴), we have developed an arrow-shooting system using compressed air as a power source, which launches the A/C/E arrow at an arbitrary velocity up to 75 m/s. We attach two points (piles) of different type (streamlined and bullet) to the arrow-nose. The boundary layer is laminar for both points for Re less than about 1.2 × 10⁴. It becomes turbulent for Re larger than 1.2 × 10⁴ and the drag coefficient increases to about 2.6, when the bullet point is attached. In the same Re range, two values of drag coefficient are found for the streamlined point, of which the lower value is about 1.6 (laminar boundary layer) and the larger value is about 2.6 (turbulent boundary layer), confirming that the point-shape has a crucial influence on the laminar to turbulent transition of the boundary layer.

Appendix

We explain in this “Appendix” the details of two data-analysis methods used to estimate the drag coefficient C _D [18]. The first is based on the ratio of the horizontal velocity components at Camera 1 and Camera 2. The second utilizes the change of arrow attitude with the gravitational acceleration g. Both methods assume that the forces acting on the arrow are the gravity and the drag, i.e., no lift is exerted on the arrow.

1.1 Method 1: using the horizontal velocity components

Under the previous assumption, the equations of motion are written as,

$$ \frac{{\rm du}}{{\rm d}t}=-\frac{1}{2}{\hat D}u\sqrt{u^2+w^2}, $$

(3)

$$ \frac{{\rm d}w}{{\rm d}t}=-g-\frac{1}{2}{\hat D}w\sqrt{u^2+w^2}, $$

(4)

$$ \frac{{\rm d}s}{{\rm d}t}=\sqrt{u^2+w^2}. $$

(5)

Here, u and w denote the horizontal and vertical velocity components, respectively. The horizontal and vertical coordinates are x and z, respectively, and s means the length along the trajectory. The velocity decay rate ${\hat D}$ (m ⁻¹) is linked with the drag coefficient C _D as

$$ C_{\rm D}=\frac{4 m {\hat D}}{\rho \pi D^2}. $$

(6)

Here, m denotes the mass of the arrow.

We can eliminate the time variable t from (3) and (5):

$$ \frac{{\rm d}u}{{\rm d}s}=-\frac{1}{2}{\hat D}u. $$

(7)

It is solved to give a simple relation:

$$ u=u_1\exp\left({-\frac{1}{2}{\hat D} s}\right). $$

(8)

Here, u ₁ is the horizontal velocity component at Camera 1. Then we have ${\hat D}$ in the following form by putting u = u ₂ with u ₂ being the horizontal velocity component at Camera 2:

$$ {\hat D}=-\frac{2}{s}\ln\left(\frac{u_2}{u_1}\right). $$

(9)

Although this relation is exact, it is not easy to determine trajectory length s between Camera 1 and Camera 2, analytically. When the initial velocity is very large, s can be approximated fairly well by the horizontal distance x between the two cameras. However, this approximation becomes too crude for a lower initial velocity, since in this case the vertical velocity component w ₁ at Camera 1 cannot be neglected. We then approximate the trajectory by a parabola assuming the drag force is weak compared with the gravity:

$$ \begin{aligned} s &= \frac{u_1^2}{2g} \left[ \left(\frac{gx}{u_1^2} - {\rm sinh\xi_1}\right) \sqrt{1 + \left( \frac{gx}{u_1^2} - {\rm sinh\xi_1} \right)^2} \right.\\ & \quad + \left. \ln \left( \frac{gx}{u_1^2}-{\rm sinh\xi_1}+ \sqrt{1 + \left( \frac{gx}{u_1^2} - {\rm sinh\xi_1} \right)^2} \right) + \frac{1}{2}{\rm sinh (2\xi_1)} + \xi_1 \right]. \end{aligned} $$

(10)

Here, a new variable $\displaystyle{w_1}/{u_1} = {\rm sinh\xi_1}$ is introduced, which is linked to the initial angle of arrow trajectory to the horizontal x-direction. We will see below that a similar variable is conveniently used in the second method of data-analysis.

1.2 Method 2: using the arrow attitude

The vertical component of the equations of motion (4) can be solved by using the variable ${w}/{u} = {\rm sinh\xi}$ or putting

$$ w=u{\rm \sinh \xi(s)}=u_1\exp\left(-\frac{1}{2}{\hat D}s\right){\rm \sinh \xi(s)}. $$

(11)

Here, we use the relation (8). After some algebra, we find that the variable ξ(s) is governed by the following differential equation:

$$ \frac{d\xi}{ds}=-\frac{g}{u_1^2}\frac{\exp({\hat D}s)}{{\rm \cosh^2\xi}}, $$

(12)

in which time t is eliminated using (5) again. This equation is easily integrated to yield the following relation:

$$ \begin{aligned} \left[{\rm \sinh 2\xi}+2\xi \right]_{\xi_1}^{\xi_2}&=-\frac{4g}{u_1^2}\frac{\exp({\hat D}s)-1}{D}\\ &=\frac{4g}{{\hat D}}\left(\frac{1}{u_1^2}-\frac{1}{u_2^2}\right). \end{aligned} $$

(13)

Since we know ξ_1,2 and u _1,2 from the video images, ${\hat D}$ is calculated using g = 9.80 m/s². Note that the second method does not require the distance between the two cameras, and that it is more convenient in outdoor measurements.