杏吧原创

Spearhead assault

Question: What is the physics involved in achieving the maximum range when
throwing the javelin?

How is it affected by the angle at which it is thrown, and what forces act on
the javelin during its flight?

Answer: The physics involved in achieving the maximum range and the forces
acting on the javelin are no different from those found when throwing any other
projectile.

The same mechanics are used for determining the flight paths of shells fired
from modern cannons, and although the designers might not have known the physics
they used similar rules of thumb for catapults, crossbows and any other
primitive missile-based weapons.

Air resistance and gravity are the primary forces working against the javelin
in flight. The air resistance is very small, because a javelin is designed to be
aerodynamic. Even so, moderate wind from the side can push the javelin
significant distances sideways鈥攋ust ask anyone who plays golf about this
phenomenon, and think of how much smaller than a javelin a golf ball is.

Gravity is working to pull the javelin back towards the ground. Although this
does not affect the javelin鈥檚 horizontal motion directly, the javelin is more
likely to land sooner if it is launched at an angle that is very close to the
ground, or conversely, an angle that is too close to vertical.

The optimum release angle for any thrown object is 45 degrees, or exactly
halfway from the horizontal to the vertical. The relatively simple geometry used
to prove this statement can be found in any fundamental text dealing with
kinematics.

The javelin rotates about its long axis as it travels through the air, a
feature that helps to keep it stable in flight鈥攋ust like the rifling in
the barrel of a gun that causes the bullet to spin. The rotation is achieved in
two ways: by the method of release from the thrower鈥檚 hand, and by designing the
javelin in such a way that air passing over it causes it to rotate.

Of course, the force behind the javelin when it is released is a prime factor
in its final distance. This is intuitively obvious, and you can find the
mathematical proof in the same text in which you read about the 45-degree throw.
Raw strength and a bit of technique, from what I understand, is all one needs
for this component.

I hope to see you at the next Olympic games.

Mike Dappolone

Hoboken, New Jersey

Answer: The forces that act on a javelin during its flight are gravity and
air resistance. The air resistance tends to slow the javelin and the gravity
tends to bring it down.

The range of the javelin depends on the initial horizontal and vertical
velocity imparted to it by the thrower. It works out this way: how far can the
javelin travel at its horizontal velocity before gravity gets the better of its
upward vertical velocity and brings it down to earth?

According to the equations of a branch of physics called projectile motion,
the range is maximum when the initial velocity imparted to it is divided equally
into vertical and horizontal velocities.

This means that under ideal conditions (in a vacuum, so there is no air
resistance) the optimum angle at which to throw it is 45 degrees. In reality,
the angle that gives maximum range depends on factors such as the wind. If there
is an opposing wind, an angle less than 45 degrees would achieve the maximum
range and vice versa.

Lakshmi Narasimhan Chakrapani

Atlanta, Georgia

Answer: The physics involved with the javelin throw deals almost exclusively
with projectile motion. There are two directions in which you are throwing the
javelin鈥攈orizontal and vertical. The basic equations are

x = v_ht

where x is the horizontal distance, v_h is the horizontal
velocity and t is the time; and

y= v_vt + 1/2 gt²

where y is the vertical distance, v_v is the vertical
velocity, t is the time and g is the acceleration due to
gravity. So when you throw the javelin with a velocity v at an angle &thgr;
to the horizontal, the equations become

x = vcos&thgr;.t

and

y = vsin&thgr;.t + 1/2 gt²

Combining these equations, you find that for any given v, the value
of x is a maximum when &thgr; is 45 degrees.

But how big is v? This is given by the impulse/momentum equation
Ft = mv (where F is the force being applied by the
thrower and m is the mass of the javelin).

To maximise the velocity, you have to apply a large force for as long a time
as possible. This is why in throwing and hitting events in sport you are often
told to follow through.

The forces acting on the javelin in the air can be minimal. The force of
gravity is, of course, the main force here. Others may include the force of the
wind, which is in turn affected by the aerodynamics of the particular javelin,
especially the smoothness of the shaft: the rougher the shaft the more air
resistance.

Jeremy Daugherty

Carlinville, Illinois

This week鈥檚 questions

Moonshift: I鈥檝e heard that the Moon is moving away from the Earth by about 4
centimetres every year. What mechanism leads to this shift and how is such a
tiny amount calculated?

Bill Gillgrass

Norwich

Light show: Why are the Northern and Southern Lights鈥攖he Earth鈥檚
auroras鈥攂oth made up of changing colours?

Guy Edwards

Chester