# What characterizes recoil?

I noticed that people who shoot 9 mm pistols on YouTube experience quite some recoil but that I don't experience any recoil when shooting my compound bow, even though the projectiles of both have similar momentum. Unfortunately, I don't own any firearms and it's been a decade since I shot one, I don't even remember which one, and I didn't ever shoot a pistol (Europe, you know ...), so I can't do a direct comparison.

My compound bow shoots a 27 g arrow at 82 m/s and a 9 mm Commonwealth standard bullet has a mass of 7.5 g. 27 g * 82 m/s / (7.5 g) = 295.2 m/s So it has the same momentum as my arrows when it travels at 295.2 m/s. Its actual speed is 370 m/s, so my arrows have 80% of the momentum of these bullets.

I already almost dismissed the question after I noticed that maybe most of the recoil comes from the exhaust gases, maybe the energy matters, not the momentum, or maybe it's because a bow spends more time accelerating its projectile than a pistol.

But then I went the other way and calculated that a 1 kg metal ball would have to travel at 2.2 m/s to have the same momentum as an arrow. If one were hit by such a ball, the impact would probably be quite noticeable, even when one catches the ball and decelerates it over the distance a bow needs to accelerate an arrow.

So my questions are:

• What characterizes recoil? Is it momentum? Is it energy? Is it something else?
• How does a compound bow manage to have no recoil?
• How are you sure that "the projectiles of both have similar momentum" ? – user139621 Oct 1 '17 at 15:48
• I calculated that the arrow has 80% of the momentum of the bullet (2nd paragraph). The values are measurements of my bow/arrow (I measured several arrows to make sure the values are consistent) and from Wikipedia (en.wikipedia.org/wiki/…). – UTF-8 Oct 1 '17 at 15:52
• This is interesting. Here's what Wiki says: "When a compound bow is drawn, the limbs are pulled in toward each other, by the buss cables, unlike a longbow or recurve where the limbs flex in the direction of the bow string. This difference allows modern compounds to have limbs that are horizontal instead of angled. The horizontal limb configuration minimizes the recoil and vibration felt by the shooter when the arrow is released." – user139621 Oct 1 '17 at 15:55
• @Blue I understand how a compound bow's arrows travel faster compared to an old-style bow (short bow, long bow, recurve bow) because more of the force the user applies is used to flex the limbs (rather than just pulling along them which doesn't store energy), but I don't understand why the bow's architecture would matter. The arrow has a mass of 27 g and travels as 82 m/s and according to my understanding, that energy or momentum or something like that has to go somewhere the other way. But there only is user holding the bow. – UTF-8 Oct 1 '17 at 16:02

• Recoil happens due to the conservation of momentum.

In detail:

When the gun is fired, the explosion pushes both the bullet and the gun in opposite directions and, as the initial momentum was zero, the total final momentum must also be zero: so the gun goes backwards with as much momentum as the bullet (and exhaust gases) go forward. The shooter holds the gun, braking its backward movement (and finally transferring this momentum to Earth via friction). Braking the gun requires applying force on it: the reaction to this force, exerted by the gun over the hand, is the recoil force.

• In a bow, the force needed to ensure the conservation of momentum is already present before the arrow is shot.

In detail:

The arm holding the bow is already exerting the force necessary to equilibrate the elastic force which, once released, propels the arrow. So when the arrow is shot, the bow exerts a decreasing force on the arrow, and the arm holding the weapon actually has to exert also a decreasing force on the bow, a sort of small "reverse recoil".

Considering, for simplification, a spring device instead of a bow (non-compound, I comment on the general case below), schematically we have, before and after the arrow is shot:

where $\bf{F}_h$ is the horizontal force exerted by the arm holding the bow, $\bf{F}_d$ the force drawing the bow (pulling it back), and $\bf{F}_e$ (shown below) the force from the spring.

Let's consider the forces when the bow is fully drawn:

The force that will propel the arrow ($\bf{F}_e$) is already being exerted by the bow, and being balanced by the archer's drawing force $\bf{F}_d$. The arm holding the bow exerts only a reaction force and, as can be seen from the free body diagrams, $F_h=F_e=F_d$.

Once the spring is released ($\bf{F}_d=0$), the spring exerts a decreasing force on the arrow and an equal (and equaling decreasing) force on the bow. That's why there's no recoil: shooting the arrow doesn't require in any moment the arm holding the bow to exert increased force, on the contrary.

Considering now a compound bow, where a system of pulleys allows $F_d<F_e$, or even a crossbow, which, after cocked, allows $F_d=0$, the recoil can remain negligible, but only because the machine now has heavy moving parts, so the recoil depends on the specific construction. Particularly important is the way the limbs move, if, e.g., they move only sideways (orthogonal to the direction of shooting), the bow receives as much momentum backwards as it imparts to the arrow forward and, just like with a gun, the shooter wasn't already exerting the force to brake the bow, so there'll be a noticeable recoil in this case.

• Yes,I think this is actually observed in reality but it's a very small effect. Watch the very first clip of this video. When you look at the riser (the part of the bow where it's held), you can observe that it moves forward. This is actually the reason some archers (see 0:52 or 1:27) tie strings around their fingers or use small devices specifically designed for this purpose so they can catch their bows without needing to encompass their risers with their fingers (which helps with precision because then the left arm only has to withstand pushing). – UTF-8 Oct 1 '17 at 17:48
• Is what you're saying that if I shoot a bow which has some final draw weight (when the bow is fully drawn), that force is exactly equal to the force the bow is pushed back with when the arrow starts accelerating? If this is the case, then: Compound bows have so-called let-off. Let-off means that the pulley system is designed in such a way that when the bow is fully drawn, the user doesn't have to bear the maximum draw weight. My bow has 65–70% let-off which means that I only have to bear 30–35% of the 65 lbs max. draw weight whilst aiming. Shouldn't I feel a recoil proportional to the let-off? – UTF-8 Oct 1 '17 at 17:56
• That's a good question, I think related to the extreme one of the crossbow - which, to the best of my knowledge also doesn't (noticeably?) display recoil. I'll think about it and elaborate in my answer. – stafusa Oct 1 '17 at 17:59
• Regarding your last paragraph: If $F_h=F_d=0$, one should experience recoil as soon as the projectile accelerates. In reality, the recoil of a modern crossbow depends on how much the limbs move forwards (It's possible to construct it in such a way that they almost only move orthogonal to the direction of fire.) and which direction they move in (They can move forwards like the limbs of medieval crossbows did or backwards.). There has to be a force counteracting the projectile's acceleration and if the user didn't feel it before shooting, they notice a difference in force when shooting. – UTF-8 Oct 2 '17 at 8:23
• Yes, it does. I'm not sure I can visualize the point where the bow "has its greatest draw weight", but the description sounds consistent. At any rate, it does seem like one of the things they try to minimize when designing bows is recoil, which makes sense. – stafusa Oct 2 '17 at 10:55

Very briefly - momentum is the result of force applied over time. A pistol applies a lot of force over a very short time - in a bow, a smaller force is applied for a much longer time.

The result is that you feel less recoil even though the momentum might be similar. F$\delta t$ is the same - but if $\delta t$ is smaller , F must be larger.

• I believe this answer is incorrect, as it states that there's (backward) recoil from shooting a simple bow, something that is not observed. See my answer. – stafusa Oct 2 '17 at 9:20
• @stafusa when you hold the bow under tension, there is force on the string and the bow. As you release the string the force on the other hand doesn't immediately disappear - but neither is it suddenly bigger. But this difference in force is exactly the net force on bow plus string plus arrow that is needed to allow the arrow to be propelled forward. Also the center of mass of the bow itself moves forward during the release, which means the force is even greater. It doesn't "feel" like recoil but it is. – Floris Oct 2 '17 at 11:52
• Perhaps it is the "re" part of the recoil where we are arguing. The force that the holding hand experiences doesn't go up when the string is released because it was already there; but if you look at the net force on the shooter (who was holding bow and string), then the shooter does experience a net backward force upon release of the string. – Floris Oct 2 '17 at 12:13
• Yes, I agree. In the moment the string is released, it stops pulling the archer forward - but the bow keeps pushing the person backwards. Net forces of same magnitude are propelling the arrow and the archer in opposite directions. But, as you noticed, I'm not calling that "recoil", as there's no increase in the force on the arm holding the weapon. – stafusa Oct 2 '17 at 14:54
• In a sense it is the arm releasing the string that recoils. – Floris Oct 2 '17 at 17:29

There are a few factors playing in to this, and not all of them physics.

Firstly the physics. The momentum of the projectile is (in the case of constant acceleration) force x time. The force is what your experience as recoil. The time over which the projectile is accelerated is very different between a bullet and a bow. For a bullet it is going to be essentially the length of the barrel divided by the speed (give or take a factor of two for the average rather than final speed, but since we are fudging things and pretending the acceleration is constant when it isn't, we're already wrong, and being over precise when making a very approximate calculation is pointless). For the bow it is the length the arrow is drawn back from the neutral position divided by the arrow speed.

The gun gives you a shorter time by both measures: the barrel length is shorter than the draw of the bow, and the final speed is much higher (by a factor of 4, using your numbers).

Additionally guns have an additional factor of the exhaust gases providing recoil to the gun but not energy to the bullet, so the momentum of the bullet tells you about the lower limit of the recoil, while a bow had no such mechanism: any inefficiencies are turned in to heat and vibration in the bow, not additional backwards momentum of the bow.

There may also be a physiological issue. Holding a gun is more or less a rest state. Holding a drawn bow, the muscles are already under tension. How the body experiences the recoil force is likely to be quite different. In a bow the arm holding the bow is pushing forward with considerable force, already braced against the recoil. With the gun the muscles that are going to oppose the recoil can be completely relaxed until the recoil impulse arrives.

• I think this answer is incorrect, as it states that there's recoil in a shooting bow, something that is not observed. – stafusa Oct 1 '17 at 17:55