Aaah I had programming mistake with the drag coefficient :smack:
I was thinking that a range of 60 meters with a composite bow was a little short :dubious:.
After re-checking again’s the article figures and re-doing the simulation, I now have an optimal arrow with a mass of 16 grams and flying 319 m (38 deg shot angle). But now the initial velocity is 86 m/s and the terminal velocity is 15 m/s. The impact is at 42 m/s.
So the arrow would be flying way faster than its terminal velocity at all times…
% Drag coeff for M = 20g
k = 2.31e-3 ;
% Initial angle
a = 38;
clf
Flight_Distance = ;Impact_Velocity_Flight = ;Terminal_Velocity_Drop= ;Initial_Velocity= ;
m_x = 10:100 ;
for M = m_x
% Simulate an arrow shot
% Initial velocity
V0 = sqrt(77^2*20/M) ;
Initial_Velocity(end+1)=V0 ;
% Simulation
Position = [0 0] ;
V = [cosd(a) sind(a)]V0 ;
i = 1 ;
dt = 0.001 ;
G = [0 -9.81] ;
while Position(i,2)>=0
i=i+1 ;
% drag
D = -V(i-1,norm(V(i-1,:))(k20/M) ;
% update elocity
V(i, = V(i-1,:)+(D+G)*dt ;
Position(i,:)=Position(i-1,:)+V(i,:)*dt ;
end
subplot(221); hold on
plot(Position(:,1),Position(:,2))
Flight_Distance(end+1) = Position(end,1) ;
Impact_Velocity_Flight(end+1) = norm(V(end,:)) ;
% Simulate an arrow drop to compute terminal velocity
% Initial velocity
V0 = 0 ;
% Simulation
Position = [0 1000] ;
V = [cosd(a) sind(a)]*V0 ;
i = 1 ;
dt = 0.01 ;
G = [0 -9.81] ;
while Position(i,2)>=0
i=i+1 ;
% drag
D = -V(i-1,:)*norm(V(i-1,:))^2*(k*20/M) ;
% update elocity
V(i,:) = V(i-1,:)+(D+G)*dt ;
Position(i,:)=Position(i-1,:)+V(i,:)*dt ;
end
subplot(222); hold on
plot(V(:,2))
Terminal_Velocity_Drop(end+1) = -V(end,2) ;
I think it’s reasonable to assume that the potential energy stored in the bow is constant regardless of the mass of the arrow (before releasing). Maybe to think of the bow as a spring (1/2kx2) where k is the spring constant and x is the pulled distance. Also an approximate assumption can be made that the same energy is transferred to the arrow.
You’ve probably already checked this, Oukile, but just to be sure, could you try your simulation with a launch angle of 90º, to make sure that in that case the final velocity is (slightly below) terminal velocity?
EDIT:
Actually, that can’t be right. The vertical component of velocity can’t be greater than terminal velocity, so if the final velocity is 42 m/s, you’re coming in at an angle of 21º or shallower. But with air resistance, the impact angle will always be steeper than the launch angle.
[Moderating]
Oh, and I also edited in the note you asked for in post 37. But I didn’t delete the mistaken material, for the sake of learning from mistakes. That work for you?
If I had a way to determine terminal velocities I would also be able to determine best launch speeds for a particular style arrow. heavier arrows carry much better in most cases as we can use denser materials to make them. I already know minimum necessary launch speeds going for higher launch speeds is where we usually loose out. In my particular case I would like to design an arrow that had a terminal velocity of about 225 fps or approx 150 mph which shouldn’t be too hard for a 260 grain arrow but I need a 200 grain arrow that will do this to get the launch speed I need.
Right. I forgot to correct the part when I computed terminal velocity. It is 58 m/s for this arrow, actually!
I think that I’ll leave aerodynamics for tonight. Although, once the mistakes are corrected, my simulations seem to match the equations in the article I linked quite well, and the authors of the article tested them with a few arrow shots. So there should be quite some truth to it.
Btw, 58m/s of terminal velocity, seems like this would hurt…
What is the “best” launch speed? Is it the one that sends the arrow the farthest? And how does terminal velocity help you find the best launch speed?
Maybe you’re thinking that an arrow’s terminal velocity is equal to its ideal launch speed. If so, that hasn’t been shown.
You’re going for maximum distance, right? If so, you want maximum initial velocity regardless of the mass of your arrow. It’s true that a heavy arrow will carry its (lower) velocity longer than a light arrow, but a light arrow with a higher V[sub]0[/sub] will go farther than the same arrow with a lower initial velocity, full stop.
Terminal velocity is a function of gravitational acceleration, mass and and aerodynamic drag. All you care about for distance is mass, aerodynamic drag and initial velocity. Terminal velocity is a red herring.
Your physicist-archer friend’s statement, “launching an arrow faster than its terminal velocity is counterproductive,” is incorrect as written. If you’re going for distance, faster is better. For a given arrow and angle, the faster arrow will always go farther. I strongly suspect the original quote has been garbled to some degree. That’s not a criticism of you, OP, but the people here could address it if we had the verbatim statement. Until we have that, I’d say your friend’s statement falls into the category of “not even wrong.”
Again, what exactly is the question here? Is it how to maximize distance for a bow of a given pull by varying the mass of the arrow?
For a given arrow and angle, a faster launch means more distance. If you keep the arrow constant but allow the angle to vary, the faster launch will go farther still than the faster launch with a fixed angle. If you’re going for distance, faster is always better.
FYI, while I was silly and didn’t check the terminal velocity and implemented a full PDE and ran it through a Monty Carlo simulation, finding for X and then calculating the Y acceleration using:
α = Angle 45 degrees with a normal distribution of ± 22.5 degrees
v = 90 m/s normal distribution ± 10 m/s
Produced this graph.
Here are the vars I used.
• m mass
• g acceleration of gravity
• C[sub]D[/sub] drag coefficient
• ρ air density
• A area of the body perpendicular to the motion
• v = dr/dt is the velocity of the arrow
• w is wind velocity (zero)
• v[sub]r[/sub] = |v − w| is the relative velocity of the body and the wind
• C[sub]L[/sub] is the coefficient of lift
• C[sub]S[/sub] is the sideways coefficient
• i[sub]t[/sub] unit vector tangent to the path
• i[sub]n[/sub] unit vector normal to the path
While posts above correctly noted that the terminal velocity of the arrow is exceeded at launch time I also do not see a hard wall in these numbers.
While I still used a numerical solution, even when accounting for yaw I don’t see a wall. I won’t paste my work because no-one wants to see a bunch of nasty double integrals. It is still just a WAG anyway.
Hopefully someone may find this interesting, or learn to check the limiting conditions before doing a lot of math
This is just one aspect and it would greatly effect impact velocity and force if we were shooting relatively straight at a target. Increasing mass exponentially increases momentum, however since the bow is a limited source of energy it decreases speed.
Plus being a limited source of energy the height of Apex will be shorter. Same as if you use a spring to launch a 1lb ball straight up or a 2lb ball straight up. Height will obviously be less on the 2 pounder unless they are so large that drag is a bigger factor than gravity on the upward launch.
It also means that on the downward part of the the trajectory the arrow would fall much faster given a similar amount of drag. Cutting off distance achieved past the Apex of flight.
If you really want to find terminal velocity for a specific arrow fire it though a chronograph at two distances.
From there you can calculate coeffecient of drag so given it’s mass you could calculate terminal velocity.
No wind tunnel needed.
I’d suspect at that weight, given length and fletchings you’re only looking at somewhere around 150fps TV or less
If you could use a dimpled shaft and maintain stiffness you might be able to reduce drag overall but even more importantly reduce drag forward of the CG thus reducing the needed size of fletching for aerodynamic stability and further reducing drag.
Though something tells me if it worked the idea would be in use already…maybe not who knows.
Empirically it’s pretty easy to show this is incorrect by looking at flight shooting records, which are split by bow category and weight.
If you have a look, you’ll notice that the records for heavier draw bows in each category are always significantly faster than lighter bows…even if it’s the same shooter for both weights. For example:
Recurve
35 lbs (15.88 kg)
Tony Osborn 407.00
50 lbs (22.7 kg)
Tony Osborn 552.66
Compound
45 lbs (20.4 kg)
Zak Arlan Reynolds 559.16
60 lbs (27.2 kg)
Zak Arlan Reynolds 688.33
Now when you’re talking the same person, on the same day, presumably with the same conditons and technique, but a +20% difference in range, it would appear that the draw weight is clearly a major factor in the range. Now in both cases the bows will have speeds way higher than terminal velocity, so this suggests that going as fast as possible even past terminal velocity is very much the productive thing to do.
Many of the classes, primitive, english longbow and modern longbow all use wood arrows and natural material fletches. The wood arrows have a much lower terminal velocity than the carbon arrows using synthetic fletching materials.
I think you’re missing a constraint. The length of the archer’s arms and the strength of his muscles will impose an absolute upper limit on the amount of energy available, and if the bow is well-designed, the energy imparted to the arrow should be consistently close to this upper limit. So you don’t have the option of “take the same arrow, but launch it at greater speed”, since the only way left to increase the speed is to decrease the mass of the arrow (since we’ve run out of additional energy to give it).
For any given arrow, the higher speed is better. For any given speed, the heavier arrow is better. But we’re not given an arrow or a speed; we’re (probably) given an energy, and so there’s a tradeoff between higher speed and more mass.
While way past the level of math any of us or doing a carbon arrow is typically much stiffer than a wooden arrow. Arrows are bending during the shot and most papers say wooden arrows are wobbling along multiple frequencies for the duration of flight.
The velocity decay rate is increased dramatically by the fetching steering the arrow towards the direction of flight. While surface roughness and laminar vs turbulent airflow do impact drag, the C[sub]D[/sub] for arrows as measured in a wind-tunnel is always lower than is observed by experiment.
Some studies actually use recurve bows because they impart a larger more consistent oscillation which reliably makes the air flow turbulent. But this is way to far into fluid dynamics for me to explain how. But the higher “spine” number or lower stiffness of a typical wood arrow compared to carbon is another reason long bows have limited range as they start to impart more and more energy into this oscillation.
Euler’s critical load is probably the easiest math wise. But once you get past laminar boundary layer the drag coefficient goes way up. A low modulus material would transition to turbulent boundary conditions at a lower energy level.
If you look at the slow speed shots in this video you will see that wobbling.
I don’t think that all carbon fiber arrows are better than all wood arrows but in general with WAG numbers they seem to be.
The carbon arrows are much stiffer with smaller diameters as well as having stiffer and thinner fletchings. There is no comparison in the distances achieved with carbon arrows and wood arrows. The foot bows that are used to send the arrows well over a mile are shooting solid carbon rods smaller than 1/8" diameter and often weigh less than 100 grains.
Even looking at the disciplines using wooden arrows and natural fletchings, poundage makes a huge difference. Here’s the American longbow records:
35 lbs (15.88 kg)
Larry Hatfield
311.10
50 lbs (22.7 kg)
Larry Hatfield
382.60
To be honest, I really can’t see any way that increasing the launch velocity to as fast as possible would ever be counter productive when you’re trying to get range, short of the arrow bursting into flame due to air friction (which seems rather unlikely) or you ripping off a fletching and screwing up the flight characteristics of the arrow (quite a lot more likely)
norm(V(i-1,:))(k20/M) ;
