Aodhan Ite an Fhithich aodhan at
Fri Mar 24 20:09:07 PST 1995

Dia duit!

This is in response to to Savian's analysis of the physics of crossbow bolts. 
For the sake of preserving bandwidth, I have not quoted the original post.

By way of credentials, let me note that I flew model rockets for several
before finding the SCA, have written computer programs to predict trajectory
and altitude for said rockets, and have spent the last 15+ years maintaining
simulation software for the Space Shuttle.  Besides, I checked my physics with
my wife, who has two degrees in physics.


Your initial assumption is in error, and you failed to take into account a
very important force: drag.

Firstly, (1) F = M * a -- that is, force = mass x acceleration (not velocity).

As for drag, (2) D = 0.5 * (CD * RHO * A * ((V/2) ** 2)
where:  D = drag, CD = drag coefficient (a dimensionless measure of
streamlining), RHO = air density,
        A = cross-sectional area of the projectile,  V = velocity

Other equations we need are: 
  "Muzzle" Velocity: (3) Vm = ((TH - D) / M) * Tt
    where: TH = thrust imparted by bow (the F above), Tt = time over which
thrust is being imparted
  Terminal Velocity: (4) Vt = Vm - (D * Tf)
    where: Tf = time of bolt's flight
  Kinetic Energy: (5) KE = 0.5 * M * (V ** 2)

In what follows, I will use the following symbols:  
  tm  = value associated with a bolt equipped with a Thistle Missile
  mb  = value associated with a bolt equipped with a Markland blunt

If I've followed this thread correctly, then the following assumptions are
  Mtm > Mmb
  Atm < Amb

If I understand the construction of these blunts, then the following should
also be correct:
  CDtm < CDmb
This says that a bolt with a Thistle Missile is aerodynamically "cleaner" than
one equipped with a Markland blunt.

Now, let us perform a thought experiment (ala Einstein), in which we are going
to use a "perfect" crossbow (TH and Tt are constant for all firings and for
any bolt) and bolts which are identical, except for the blunts. 

Looking at equation (2), we see that, for a given V, Dtm < Dmb, because CDtm <
CDmb and Atm < Amb.  This implies in equation (3) that a Thistle Missile loses
less of the thrust of the bow to drag than does a Markland blunt.  However,
note the feedback loop here.  Less drag implies greater velocity, which
increases drag in proportion to the square of the velocity.  Without actual
numbers, it's hard to predict which bolt will have the greater Vm, based on
their aerodynamics.

However, there is another factor involved here, the respective masses of the
bolts.  Mtm > Mmb and TH is inversely proportional to mass.  The difference in
masses is probably more important that the differences in drag, so <
Vm.mb (maybe).  Again, without some real numbers, it's hard to say what will

Equation (4) has another feedback in it.  The slower bolt has less drag per
equation (2) and so loses less of its initial velocity over a given time
period.  But, it takes longer to get there, so there's more time for drag to
slow it down.  Which bolt has the greater terminal velocity?

Equation (5) tells us that the more massive bolt has more energy for a given
velocity and the the faster bolt has more energy for a given mass.  Which of
our two bolts has more energy depends on the ratios of the masses and the

I'm not trying to say Savian's analysis is wrong, per se, but that the
situation too complex to say anything without some hard numbers.  Give me a
wind tunnel to measure the CD's and a scale to measure the masses, and then we
can make predictions.  Or, get some flight times so we can have velocities to
use with the masses to determine KE.

Savian is correct in stating that the stiffer Thistle Missile will have a
harder impact.  This is because less of its Kinetic Energy is spent
compressing the blunt, thus imparting more energy to the target.

I also agree with Savian that it is hard believe reports of excessive force
from crossbow bolts without seeing the dents they caused.  I've seen plenty of
helm-denting sword blows called light (may have even called a few that myself,
for all I know...).

Feicfidh me' ari's thu',

Baron Aodhan Ite an Fhithich    aodhan at
Master of the Laurel            Lough na Dobharchu' BBS  1-713-338-2570
Dobharchu' Herald               "Your Information Roman Road"
mka David H. Brummel            1:106/22  180:11/22  762:2200/2
SCA Member 02245                Barony of Loch Soilleir, Ansteorra
plummety sable and argent on a sun of eight rays or a feather bendwise
sinister sable
No bird soars too high, if he soars with his own wings. - Blake

More information about the Ansteorra mailing list