Whatever the outcome of the previous section, we'll take it to give us an initial velocity for the arrow as it leaves the bow. Together with it's angle of departure this ought to be enough to predict the arrow's trajectory from launch to impact.

During it's flight, the weather will continuously influence the arrow. Sudden winds are a constant nuisance during outdoor target practise, and gusts of rain are easily worse. In this simple model I'll assume a rigid arrow, always travelling in the direction it's pointing. There'll be gravity and air friction, but nothing else.

(Of course wind and weather influences can be taken care of, but I'm already using drag-formula's that someone else found to be working. And have neither knowledge of how sidewinds or individual raindrops actually influence an arrow nor do I want to know.)

Gravity being only vertical, it seems natural to split the arrows velocity in a horizontal and a vertical part. The model can be made like this:

horizontal : Mass x hor. acceleration = - hor. component of drag vertical, ascending : Mass x vert. acc. = - gravity - vert. comp. of drag vertical, descending : Mass x vert. acc. = - gravity + vert. comp. of drag

We know the arrow's mass, expect gravity to be constant, and now must find an expression for the drag force pulling at the arrow. The drag formula has to be built up out of individual arrow parts. Reynolds numbers to predict air flow are calculated at typical arrow speeds ranging from 30 to 100 meters/second. - Both the tip and the nock drag components are described the same way:

tip/nock drag force = tip/nock coefficient * facing surface * ½ air density * velocity squared

The coefficients vary with the shape of the tip/nock, but for typical nocking the coefficient is 0.30 while the so called bullet tip has coefficient 0.05.

the facing area is in most cases just

p * shaft radiussquared.

- The shaft Reynolds number is of the order of millions and it is therefore treated as a turbulent flow problem. There has been done some research on the drag function's shape [see reference], but here I'll merely state it to consist of
( c1 * Re^-0.2 + c2 * shaft length / shaft radius * Re^-0.4 ) * shaft surface * ½ air density * velocity squared

c1 and c2 are two constant of order 0.074 and 0.008, The Reynolds number Re is given byRe = air density / air viscosity * velocity * typical length

.

Seeing that velocity appears in the Reynolds number, for a typical arrow we can reduce this to (v stands for velocity, c3,c4 constants):

shaft drag force = c3 x v^1.8 + c4 x v^1.6

- The fletching Reynolds number is on the edge of laminar/turbulent flow but seems to be described laminar. My source gives:

fletching drag force = c5 * Re^-0.5 * all fletching surface * ½ air density * velocity squared

This can be reduced to a simple shape in the same way:

fletching drag force = c6 * v^1.5

Adding everything yields the following drag force function:

drag force = d0 * ( d1 * v^2 + d2 * v^1.8 + d3 * v^1.6 + d4 * v^1.5 )

For a typical arrow at say 50 meters/second d1,d2,d3 and d4 are respectively 0.17 , 0.25 , 0.52 and 0.06 , which shows that shaft drag amounts 77 % of the total drag. While the fletching seems to be doing surprisingly little.

In some books this is put together to

drag force = d6 * v^1.8, which is a good approximation for the usual arrow speeds. ( For a 30" standard arrow with mass 0.0021 kg as used on a 40 lb. recurved bow, d6 = 0.0070 .)

For our model this does not matter very much because the two coupled differential equations won't be analytically solvable anyway. We'll have to resort to numerical methods.

I found no trouble in doing so, but have not yet figured out how to present the results in ASCII.

For what I think my bow does, the 90 meters trajectory was found to be like this: initial velocity 56 m/s, launch angle 11 degrees. Arrow reaches max. height of 6.70 m after 0.96 s at 47 m before plunging into the target after 2.02 s at a speed of 35 m/s.

So for the 90 m distance this says my arrow looses 40 % of his speed and that about 60 % of it's energy was consumed by drag.

I took the friction description from the following book:

'Physical laws of archery' '88 (fourth ed.'91) by T.L. ListonFor those who want to calculate some trajectories without having to program it themselves, I included the crude pascal program I used. (Only in Dutch at the moment, I'm afraid.)

As said before a real arrow isn't rigid at all. At the launch, it is forced to accelerate so fast that it buckles under the strain. It is said that aluminium shafts are better to use on short distances because they are faster to reach a stable flight. The carbon-based arrows weight about the same, but are thinner, have far less surface, thus experience less drag and are faster to reach the target. nevertheless, they are said to need longer to reach a stable flight, and are therefore thought to be less suitable for shorter distances.

Most archers (I know) find it too much trouble to tune and re-tune their bows between the longer and shorter distances and - if they can afford them - use carbon for whatever distance they're practising.

The fletching is placed so to make the arrow spin about it's length axis, thereby levelling out faults due to slightly bent shafts or unbalanced fletching and such.

Whether the mechanical energy stored in an arrow - buckled as it is after departing the string - comes to some good use by rowing in the air using the fletching or something I wouldn't know. This paragraph is merely to show the reader there's a lot more of thorny things in the bush, and that the previous section may once again provide to simple a model.