By recording it’s speed at two points you could calculate it’s coeffecient of drag in that velocity range, then chart it’s increase in drag with relation to increases in speed. Then map that into trajectories to figure out the point of diminishing return but it’s unlikely to have any direct relationship with terminal velocity.
By using a lighter arrow you aren’t sacrificing terminal velocity (strictly speaking you’d want that as slow as possible, since that has only to do with falling) so much as momentum.
Momentum does increase drastically with an increase in mass compared to increasing speed. Problem is the bow has limited energy to give and on the downside it falls faster given similar drag.
So your balancing act is where it’s at.
Wow.
Thanks
I am having a hard time imagining an arrow shaft that light. I guess with a really light wood like cedar it is possible.
What is the diameter of such a light arrow? (I assumed 1/4in diameter but I have no idea)
I certainly am rusty with my mechanics, but it seems to me that you would want as heavy an arrow as possible so that for a given bow the velocity would be low and momentum high. With reduced air resistance the momentum will carry the arrow further. I believe one needs a 45 deg angle to maximize the time of flight but that is true of every arrow weight and bow strength.
Drag on things is exponential with speed increasing the exponent…
I think when he said “terminal velocity” he really mean “the fastest you shoot an arrow”… trying to go much faster causes the speed to wipe off in short range, AND the arrow to bend and wobble and go off course. (And other things such as misbehaviour in the bow.)
terminal velocity is just where the drag force is the same as its weight force.
As to the side track of actually a stone dropping,only ever approaching terminal velocity. I think it will reach terminal velocity, as it would get very close to it, if the atmosphere was perfectly homogenous… But the actual atmosphere has density changes in the various layers… as the air becomes more dense generally near the ground, terminal velocity becomes slower… so it does drop into a situation where its speed is exceeding terminal velocity at that location. The variation in terminal velocity (measured instantaneously) just due to the weather (heat,wind,humidity,etc) is also large enough to cause same thing.
We actually try and find very dense heavy but stiff woods to make our arrows out of. The best flight arrows are made from tonkin can which is almost double the weight of the heaviest woods. I am shooting in the 50# weight class. Small diameters under 1/4" for my arrows is desirable. Below 1/4" stiffness becomes an issue. I have some wood right now I have been using where I am able to hit about 3/16 diameter but a little heavier than I like.
The more I think about it I am starting to think that his reference to terminal velocity has more to do with the fact that an arrow going much faster than that will loose the speed so quickly that opting for more weight and less drag is a better option.
Are you allowed to use laminate arrows? Because that would open up a whole new option in making the arrows with the desired stiffness/flex.
I am confused by some of these responses.
Terminal velocity is when the force of gravity balances wind resistance. As such, it only applies to the vertical component of an objects flight. If you drop an object out of a helicopter or off the leaning tower of Pisa, it will accelerate due to gravity until the force of wind resistance balances the force of gravity on the object.
For the horizontal component of an arrow shot form a bow, the terminal velocity is zero. It is given an initial boost by the bow, and from then on, air resistance slows the horizontal component in some way, some exponent, proportional to its velocity and the cross section of the arrow. (and add secondary effects like drag due to oscillation of the shaft due to excessive force from the bow) So the heavier the arrow for the same cross section then the further it goes, as resistance force bleeds off momentum.
You can shoot an arrow as hard as you want to, it will rise up slowed by gravity and wind resistance, and eventually reach the apex of its flight. The harder it is fired, the higher it goes. The ultimate trajectory limit will be the point where the arrow goes so high, that forward velocity has drained off completely and the arrow finishes its trajectory plunging … wait for it… straight down at terminal velocity. In that case, professor Proton is right.
However, it’s logical to see than if an arrow is propelled from a bow faster than terminal velocity, it is entirely possible for it to hit a target not too distant before wind resistance has dropped its speed down below vertical terminal velocity. But in the more logical concept of arrow target practice, where the target is hit on the downward arc of the trajectory from a fairly decent distance - odds are the remaining trajectory is close to terminal velocity.
The record for an arrow shot for distance is over 900 feet. Beyond this, I assume a hand-drawn bow is incapable of propelling an arrow fast enough that the arc reaches high enough - i.e. a 45 degree approximate trajectory means that at the apex is after about 450 feet (a little more due to ongoing wind resistance horizontally).
Model rockets do go further, higher (although we tend not to aim them at a 45-degree angle) even though they have worse wind resistance than arrows.
So the answer is - the faster, the better.
It’s not directly terminal velocity which is relevant. But there’s some speed above which drag is overwhelming, above which increasing speed will suffer from serious diminishing returns. And terminal velocity is a good back-of-the-envelope estimate for what that speed would be.
To know exactly what the optimum speed would be, we’d need to know exactly what the tradeoffs are to getting more speed. If we’re given the choice of the same arrow and any speed we’d like, in still air, then it’d always be best to give it as much speed as possible. But if, for instance, we have a limited amount of energy available, and we’re varying the speed by varying the mass of the arrow, then a lighter (i.e., initially faster) arrow is going to lose more energy.
One of our shooters a scientist and engineer uses laminated tonkin cane arrows constructed the same way tonkin cane flyrods are built. he has seen great success with them as his arrows have been used to set the majority of records in the classes that use natural material arrows.
It actually does make more sense taken in this context. We still have the problem of trying to figure out an individual arrows terminal velocity. I wonder if it could be tested in fluids like water? Very short distances would be required.
An upscaled version in a wind tunnel would probably be easier and avoid the need to make a really high speed wind tunnel. The problem with water is it would be hard to figure out what happens when the flow is no longer laminar.
A small wind tunnel for wood arrow testing would need no more than 200 mph wind I suspect. I was going to suspend an arrow from a string and see how much wind it could take before it started putting slack in the string. I am not so sure I could keep an arrow straight enough to test like this unless I had it suspended from both ends.
Fletching is usually set at an angle to impart spin to improve accuracy, you would need to find the drag with axial spin and at various angles of attack.
Remember the force vector for gravity is also exclusively perpendicular to the shaft. While the drag in the forward direction will have an impact it is the horizontal drag that would be the limiting factor on vertical speed.
The relative wind will be the important factor.
I tried a simple numerical integration model to check this situation, and here are my results:
Assuming Air resistance is proportional to the square of velocity, we get two equations (y is the vertical direction) :
(dVy/dt) = -g -A(Vx^2 + Vy^2)*sinΘ [where Θ = tan_inv (Vy/Vx), and A is a constant]
(dVx/dt) = -A(Vx^2 + Vy^2)*cosΘ
Assumed Vinitial = 125 m/s and all SI units
Here is the interesting thing : The value of A determines the optimal angle of shooting. I did an order of magnitude check and for A =0.02 and the optimal angle works out to be 30 degrees or less for the longest range.
Given that HoneyBadgerDC gave an optimal angle of 38 to 45 degrees, the order of magnitude value of A works out to be 0.001 (SI units).
With these numbers, the arrow ** never reaches ** terminal velocity. The acceleration does reach low numbers (0.3 m/s2 on the x and .9 m/s2 on the y) but its not 0.
Since distance is the goal it would seem you need to be concerned with the ballistic coefficient just like we do with bullets. A quick search shows that arrow BC have been discussed in numerous forums and web sites. For a given arrow diameter, weight and length the point becomes the easiest thing to alter. Are there different points available or have they pretty much been optimized over the years? Just guessing I think a medium length ogive would be good.
Another thing arises from my model rocketry experience and that is the degree of stability. You want the fletches to give stability but not too much stability. An over stabilized projectile corrects too quickly and produces a “jagged” flight path.
I am not an archer but probably these things are the topic of many a bull session among archers.
Dennis
I am always captivated by HoneyBadger’s flying archery posts, and this time I went for a bit of digging and numerical simulations.
First, I found some considerations and on arrow physics here. They made some measurement with a composite bow and some standard arrows.
From there, I got a drag equation:
dV/dt = -k.|V|^2
With k = 2.31e-3 for a reference arrow of 20g. Changing arrow mass M while maintaining the same aerodynamic profile will result in a k decreasing proportionally to M (since the drag force F will remain constant and applying M.dV/dt = F):
k = 2.31e-3.20/M
Next, I got an initial velocity of V0=77 m/s for the same arrow and the bow used in this study. Assuming that the bow provides the same kinetic energy to each shot (M.V0^2 = 20.77^2), then initial velocity is inversely proportional to the square of arrow mass (btw it is fine to write M in grams in this case).
V0 = sqrt(77^2.20/M) = 77.sqrt(20/M)
With and given the initial angle, I can simulate the arrow’s trajectory and flight distance. I can also simulate a drop from a height of say 1 km to compute its terminal velocity.
So, I found:
[numerical mistake; see post #41 below for corrected results]
- That a reference 20g arrow has a terminal velocity of 16.19 m/s. That sounds a bit slow but some googling found values in the same order of magnitude. This is much lower than the initial velocity of 77m/s measured in the study.
- The optimal arrow has a weight of 90g. It will fly 64m, with an initial velocity of 36.3 m/s. This arrow has a terminal velocity of 26.7 m/s. At the end of the optimal flight, it will impact at 20.5 m/s.
- In contrast, the 20 g arrow flies 36.81 m.
- An 131g arrow will have an identical initial and terminal velocities (30.3 m/s) and fly 60m.
- By the way, all this is for an optimal angle of 41 deg. Shooting at 45 deg gives a slightly lower distance of 63.9 m with a 92 grams arrow.
I imagine that these parameter differ a lot from HoneyBadger’s flight archery bows and arrows, but here is a funny thing. In this simulation, the optimal shot velocity is about 1.36 times the terminal velocity. I made a few variations, multiplying or dividing the arrow’s initial kinetic energy or drag coefficient by 4, and I found that this ratio is always about the same. So that should actually generalize to a wide range of bows.
So, the simulations confirm that there is an ideal mass, as expected intuitively: heavy arrows will shoot slowly and gravity will pull them back to earth faster; light arrow will shoot fast but drag will be relatively higher.
In addition, they actually kind of agree with your friend: there is a relation between an arrow’s optimal initial velocity and terminal velocity, with (in these simulations), a coefficient of about 1.35. Keeping in mind that these are very simple simulations, it is interesting…
Here is a Matlab code for those who are interested and can run it (in spoiler to save space):
% Drag coeff for m = 20g
k = 2.31e-3 ;
% Initial angle
a = 41 ;
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,:))^2(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)
Terminal_Velocity_Drop(end+1) = -V(end,2) ;
end
subplot(223)
plot(m_x, Flight_Distance) ;[m,k]=max(Flight_Distance);
fprintf(1,'Optimal arrow: mass = %.0f grams, distance = %.1f m
‘,m_x(k),m)
subplot(224)
plot(m_x, [Initial_Velocity; Terminal_Velocity_Drop; Impact_Velocity_Flight]’) ;
fprintf(1,'Initial velocity = %.1f m/s, terminal velocity = %.1f m/s, impact = %.1f m/s
'…
,Initial_Velocity(k),Terminal_Velocity_Drop(k),Impact_Velocity_Flight(k))
Let’s say you build a wind tunnel and accurately determine your arrow’s terminal velocity. What are you going to do with that information?
As far as I can tell (and I really might be wrong) this would be a straightforward external ballistics problem that has no direct connection to terminal velocity. I say “would” because the question isn’t clearly defined.
OP, can you please state your exact question here?
This is probably less serious than imagined. The question is how heavy an arrowhead? After all the front cross section of a bullet is about the same except for the fletching, and bullets don’t suffer too much from air resistance - but of course have a pretty good initial velocity.
The relevant equation if we’re insisting it be an arrow shot from a bow is v=sqrt(ad) presumably the best you can get from a bow is d=4 ft. Draw.
But a = F/m, and while F is limited by human muscles, m can vary for different arrows.