The following example of how to calculate an arrow (rotational drag force) centre of pressure makes two basic simplifications. Firstly the arrow is assumed to be initially stationary i.e.no linear motion, not rotating (fishtailing) and not vibrating. Secondly the shape of the fletchings is assumed to be a right angled triangle, the nearest simple shape that approximates to a fletching. The centre of pressure calculated is that relating to arrow rotation only. The drag force component acting through the centre of gravity which acts only to move the arrow (from part of the shaft) is ignored. Note that we get both linear and angular acceleration so the arrow centre of mass has to move.
The approach is first to calculate the total pressure and centre of
pressure locations for the shaft drag, fletching drag and Munk moment individually
and then calculate the arrow total pressure and centre of pressure location
from the three components.
In the diagram the total rottional drag force (T) acts at the arrow centre of pressure at distance (Lt) behind the centre of gravity (C). The total drag is the sum of the fletching drag (Ff), the appropriate shaft drag (Fs) and the Munk moment (Fm). The distances from the centre of gravity to the fletching drag, shaft drag and Munk moment centres of pressure are Lf, Ls and Lm respectively. The resulting instantaneous axis of rotatation of the arrow (R) is at a distance 'A' in front of the arrow centre of gravity.
If you equate the moments for the arrow total pressure/centre of pressure to the indivual shaft/fletchings/Munk moment total pressures/centres of pressure you end up with the distance Lt being given by:-
Lt = (Fs Ls + Ff Lf - Fm Lm)/(Fs+Ff-Fm)
Suppose the arrow is 80 cm long and 0.5 cm diameter. The arrow FOC is 13%. The fletchings are assumed to be 2.5 cm long with each fletching having a total area of 3.5 square cm. The front of the fletchings are fitted 73 cm from the front of the shaft. There are 3 fletchings fitted at 120 degrees to the shaft and they are assumed to be triangular in shape.
To make life simple it is assumed that the drag properties of the shaft and fletchings are identical so instead of drag forces we can use drag areas. (main difference is in the shaft and fletching drag coefficients).
With an FOC of 13% the centre of gravity is 10.4 cm in front of the centre of the shaft.
The value of Fm is assumed at 1.5 square cms. (based on some rough measurements made some years ago).The centre of pressure is assumed to act at the back of the arrow so Lm = 80/2 + 10.4 = 54 cms.
The shaft area contributing to rotational drag Fs = 2 x 10.4 x 0.5 = 10.4 square cms. (see section on FOC). Because of symmetry the distance Ls is half the shaft length i.e. 40 cms
For a triangular fletching the centre of pressure horizontal position is 2/3 distance along the base from the front. (The vertical position doesn't matter as the fletchings are assumed not to be spinning the arrow)
i.e. Lf = 73 + 2.5 x 2 / 3 - (40-10.4) = 45.07 cms
The effective area Ff of the fletchings = 1.5 x 3.5 = 5.25 square cms.
(the multiplier 1.5 allows for the fletching angle = 2 x sin squared (alpha/2) where alpha is the angle between fletchings).
Lt is therefore (10.4 x 40 + 5.25 x 45.07 - 1.5 x 54)/(10.4 + 5.25 - 1.5) = 40.4 cms
i.e. the centre of pressure is 40.4 centimetres behind the arrow centre of gravity.
Any arrow rotation will have the effect of moving the centre of pressure further back so the 40.4 cms represents the minimum distance between the centre of gravity and the centre of pressure. (With e.g. fishtailing the influence of the shaft drag decreases and the influence of the fletching drag and Munk moment increases)
If 'Ig' is the arrow moment of inertia at the centre of gravity and 'M' is the total arrow mass then the distance of the instantaneous arrow axis of rotation in front of the centre of gravity 'A' is given by:-
A = Ig / (Lt x M)
e.g. if Ig was 5400 gm cm squ. and M was 18 gms then A = 5400/(40.4 x 18) = 7.4 cms
In reality over time the arrow spin angular momentum varies (as it fishtails/porpoises) and its linear speed (ignored above) varies. As such an arrow does not have an "axis of rotation" (for example it's travelling at around 200fps :) ). Engineers will define the (spin) axis of rotation as being at the centre of mass but this is really a practical convenience as it decouples the linear and angular momentum. One could just as well define the axis of rotation in this sense as being at the tip of the archer's nose (not so useful though!). Some further comments on this topic here Arrow rotation
.Archers from what I've heard/read assign a different meaning to "arrow axis of rotation" which is based on how they see the arrow behaving. So they describe the arrow as "rotating at the point", "rotating in front" or "rotating behind" the point. What they are describing is the combined effect of the arrow rotation and the arrow linear movement (sideways/up-down). In this case the centre of pressure position, as regards arrow response, is best regarded as being based on the total drag on the arrow combined with (the relatively small) existing spin angular momentum. If you balance an arrow on one finger and give an upward push at various points on the shaft behind the cog then this well illustrates the overall arrow behaviour in response to varying position of the overall centre of pressure.
Last Revision 1 July 2009