BOW EFFICIENCY AND THE CONCEPT OF VIRTUAL MASS

Defining Virtual Mass

Starting with a stationary bow at the brace height position, as you draw it back you store energy in the bow. The area under the force draw curve equals the total energy stored (E_t). When the arrow is released the bow ends up stationary again at the bracing height. The total energy stored has gone somewhere. Most of this energy ends up where you want it as arrow linear kinetic energy (E_a) but the rest (E_w)is wasted.

The energy equation for the bow is thus:
E_t = E_a + E_w

The energy efficiency of the bow (F) is defined as the ratio of the arrow energy to the total stored energy. i.e.
F = E_a/E_t

Supposing the arrow leaves the bow with a particular speed (S) then you can write the total stored energy in the bow as being equal to some imaginary mass (M) travelling at the same speed as the arrow. i.e.
E_t = MS²/2

The value "M" includes the mass of the arrow (m) with the remaining mass (v) called the Virtual Mass. i.e.
M = m + v

This gives you that E_t = MS²/2 = (m + v)S²/2 = mS²/2 + vS²/2
mS²/2 is just the kinetic energy of the arrow and vS²/2 is the wasted energy E_w in tems of the kinetic energy of an imaginary Virtual Mass.

The bow energy efficiency F = E_a/E_t thus becomes:
F = (mS²/2)/((m + v)S²/2) = m/(m + v)

The bow energy efficiency can be defined in terms of the arrow mass and the value of the virtual mass. What makes this useful is that it is found by experiment that for a given bow the value of the virtual mass is a constant over a sensible range of arrow mass.

Note that this expression for bow energy efficiency and the value of v being constant indicates that the heavier the arrow then the more energy efficient the bow becomes. We'll look at the practical effects of this later on.

Finding the Value of the Virtual Mass

In order to see how varying arrow weight effects bow efficiency or arrow speed we need to know value of the (constant) virtual mass for the bow. There are a number of ways you can do this.

The first method (maybe not recommended) is just to make a guess. Let's say a typical recurve bow has an energy efficiency between 70% and 80% and a typical arrow weight is 300 grains. Using the energy efficiency equation we have now got we get:

with 70% efficiency 0.7 = 300/(300+v) which gives a value of v = 128 grains

With 80% efficiency 0.8 = 300/(300+v) which gives a value of v = 75 grains

So lets take a typical value of virtual mass at somewhere around 100 grains

Probably the easiest method to measure the virtual mass reasonably accurately is shoot arrows of different weights and use a chrono to estimate the respective arrow speeds. Suppose we have two arrows of weight m1 and m2 and measure their respective speeds out of the bow at S1 and S2. The using the equation for total bow energy above we get:

(m1 + v)S1²/2 = (m2 + v)S2²/2

or tidying things up (m1+v)/(m2+v) = S2²/S1²

From which you can calculate the value of v

The third method that can be used, though a bit cumbersome, is to measure the speed out of the bow of an arrow of known weight and estimate the total energy stored in the bow from the force draw curve. You can then use the above equation to get the value of v:

E_t = (m + v)S²/2

To get the value for the total energy you can either plot a draw weight - draw length curve on graph paper and measure the area under it (A), from which
E_t = gA or assume the draw force curve is a straight line which gives E_t = gLW/2.

Where "L" is the draw length, "W" the draw weight and "g" the gravitational acceleration. Don't forget to include "g". In archery, energy is often quoted in foot-lbs which is not a unit of energy. To convert you need to multiply by "g".

Example of Estimating Arrow Speed

Suppose you have a 300 grain arrow (m1) and a measured speed of 210 feet per second (S1). What speed (S2) would you get with an increased pile weight say giving an overall 330 grain arrow (m2). Let's say the virtual mass value for the bow is 100 grains

If you ignore the effect of arrow mass on bow efficiency then the two arrows would have the same kinetic energy i.e.
S2 = S1*Squareroot(m1/m2) = 210*Sqrt(300/330) = 200 fps

With the 300 grain arrow the bow energy efficiency = 300/(300+100) = 0.75 (75%)

With a 330 grain arrow the bow energy efficiency = 330/(330+100) = 0.77 (77%)
so the extra arrow weight increases the bow efficiency by around 2%

The speed of the heavier arrow would be 210*Sqrt((300+100)/(330+100)) = 203 fps

So the speed effect of the heavier arrow on bow efficiency is in this case around 3 feet per second

How Well Does The Virtual Mass Approach Work

To test whether the virtual mass approach works in predicting speeds for arrows of different masses clearly we need to use the method to predict arrow speeds and compare the prediction with actually measured speeds. Bertil Olssen on his excellent site has produced numbers we can be confident about so we can make the comparison. The following table lists measured different arrow speeds for different mass arrows and compares the results with speeds calculated using the above virtual mass method.The first two arrows in the table are used for the virtual mass calculation.

As can be seen in this case when more than doubling the arrow mass the estimated speed is withing 0.25 m/s

Why Does the Virtual Mass Approach Work and What are Its Limitations

The basic idea behind the virtual mass concept is that there is a linear relationship between the velocity of any moving part of the bow and the arrow during the power stroke.

If the moving parts of the bow (limbs, string etc.) are broken down into "i" elements each with a mass M_i and speed V_i and the arrow mass and speed are M_a and V_a then any V_i = K_i*V_a
where K_i is a constant. For example the part of the string in contact with the nock could be said to be travelling at the same speed as the arrow so K = 1. As you go up the string from the nock because of the geometry a part of the string would be travelling slower than the nock; K = 0.9 say.

The kinetic energy of any bow/string etc element E_i is given by 0.5*M_i*V_i².

The total kineetic energy "E" of the bow arrow system (arrow energy plus bow energy) is given by:

E = 0.5*M_a*V_a² + ∑_i 0.5*M_i*V_i²

E = 0.5*M_a*V_a² + ∑_i 0.5*M_i*K_i²*V_a²

E = 0.5*V_a² (M_a + ∑_iM_i*K_i²)

It follows that the bow virtual mass is given by ∑_iM_i*K_i².

There are two areas where the virtual mass approximation goes wrong:
Firstly there are other energy losses than KE ones e.g.the Drag force on the bow limbs and string (proportional to the velocity squared).
Secondly the values of the K_i are not constant. With different weight arrows the geometry of the limbs and string will vary; the limbs will flex in a slightly different way and the string will stretch different amounts.

Last Revision 12 June 2014