F=MA
A 550 gr arrow with 8% FOC will penetrate the exact same as a 550 gr arrow with 22% FOC.
Adding FOC is just a shortcut past properly tuning your arrow/components to your bow.
Completely incorrect.
The same amount of force is available, but the force is not equally transmitted between the two arrow setups. The arrow is not rigid. It blows my mind that almost every archer who has tuned a bow/arrow is fully aware that the arrow is not rigid at launch, yet most of them assume the arrow which was flexible at launch has magically become rigid upon impact.
Although nothing is actually rigid, we can sort of consider the broadhead, adapter, and knock as being rigid, but the arrow shaft is absolutely not rigid. Even if we loosely consider some of the components to be rigid, we need to consider them separately from each other because force much be transmitted through them.
On launch, the string applies force to the knock, and the knock, along with every other piece of the arrow applies and equal force to the string. Those forces are equal at the knock/string interface, but they are not equal throughout the bow or throughout the arrow. On launch, the knock is subjected to the highest force. Broken nocks can easily be a result of arrow mass and FOC(FOC matters because it affects bending and bending is related to leverage or torque). Every other piece of the arrow is subjected to less force than the knock. The center of the arrow shaft is being pushed from the piece of arrow shaft immediately rearward if center and the F that piece is subjected to is only equal to M of the arrow forward that point. It is not subjected to the entire arrow’s mass. Anyway, on launch, the knock transmits the force of the string to the arrow shaft, and the arrow shaft transmits it to the insert, and the insert transmits it to the broadhead. And guess what? That force bends the arrow shaft.
The shaft bends because the knock is temporarily accelerating faster than the broadhead. Now think of this in reverse. On impact, the broadhead begins to decelerate due to hitting the target. F=MA. But the broadhead and insert and not actually rigid, and there connect must definitely be dealt with when considering strength. The F
HERE is generated by the M of the insert, arrow shaft, fletching and knock being decelerated
BY THE SLOWING BROAD HEAD. Inserts do not break at impact due to heavy bread heads. Heavy broad heads, all else being equal, decelerate less upon impact. It’s tempting to argue that the broad head is pushed by the rest of the arrow and thus with equal total arrow weight the broad head will decelerate equally regardless of the broad head’s weight. That’s incorrect. The broad head decelerates first before the insert can apply a force to it, then the insert decelerates before the shaft can apply a force, then the shaft must decelerate before the fletching and knock can apply a force. It’s easy to tell yourself this cannot be because it’s all firmly connected,
BUT NOTHING IS ACTUALLY RIGID! In fact, the very tip of the broad head slows down before the portion of the broad head just rear of the tip slows down.
It all deforms and bends and vibrates to different degrees. This can be easily demonstrated with a long loosely coiled spring slightly stronger than a Slinky. Push the spring along a table into a wall. Does the spring uniformly compress throughout its length when it contacts the wall?
NO! First the portion of the spring that hits the wall compresses, and does so more than necessary and a compression wave travels the length of the spring until it achieves equilibrium. A Slinky will demonstrate an identical effect in tension.
The arrow does the exact same thing on LAUNCH and on IMPACT! No one argues on launch. There are only two differences between the arrow impacting a target and a long spring impacting a wall. A) the stiffness of the arrow is high enough that the bending and vibration is difficult to see, and B) the arrow is not a single uniform material(broad head, insert, shaft etc. all with different stiffnesses). With all of that in mind, you should now somewhat understand that the F and the A in F=MA are not the total F and A, but must rather be dealt with in pieces. To do it perfectly you would have to break it down into infinitely small pieces. If you took calculus you might remember the first week or two of drawing smaller and smaller bars under a curve, and then solving for the limit as the bars became infinitely smaller. You have to do the same thing for the arrow. You can go grab your high school or college physics text book and find a problem when a whole car gets represented by a dot and think you’ve proven me wrong, but you have not. That’s a simplification for the purpose of teaching concepts. It’s not real life. If it was real life the crumple zones and air bags would not work. Instead, crumple zones allow the driver’s seat to decelerate at a lower rate than the bumper, which decreases the peak force applied to the driver’s back by the seat, and airbags allow the drivers face to decelerate at a lower rate than the dashboard which decrease the peak force applied to the driver’s face. If the car was rigid, these to safety features would not work. Luckily, the safety features do work, and they work because the deceleration of the DRIVER is not solely dependent upon the total mass and velocity of his vehicle. The reason it is not, is because he is not rigidly attached to the car, and the car is not rigid.
If the arrow was rigid, then FOC would not matter. BUT THE ARROW IS NOT RIGID, AND NEITHER ARE RHE CONNECTIONS OF ITS COMPONENTS!!!
Now, if we wanted to calculate penetration, which to my knowledge no one can actually do with a mathematically derived equation(but somewhat can with experimentally derived equations) F=MA where M is total arrow mass would give you a theoretical maximum F to be applied to the target via the broad head in the direction of flight. Forces applied at any other angle only contribute a component to penetrating in the direction we want to penetrate. That theoretical maximum however could never actually be reached.
IF an arrow were ever to impact a target perfectly orthogonally to its direction of flight, the arrow would vibrate in a purely compression wave with the arrow shaft rapidly getting larger and smaller in a diameter. In reality however, we can’t ever get things aligned so perfectly. What happens instead is that the arrow shaft vibrates side to side. This side to side vibration causes drag on the target as the arrow penetrates the target, and the force
OF THE ARROW SHAFT, FLETCHING, AND KNOCK are what break inserts. Heavy broad heads do not break inserts. Rapid deceleration of the broad head compared to the arrow shaft, and heavy fletching and knocks are what breaks inserts. With equal design, a heavier broad will decelerate more slowly and the forces generated by the arrow shaft against the insert will be lower. However, broad head weight isn’t the only thing effect it’s deceleration upon impact. Dull broad heads, wide broad heads, multi-blades broadheads etc will all increase the deceleration of the broad head upon impact compared to a sharper, single-blade narrow broad head, which will all increase the vibration of the arrow shaft, and will decrease penetration more than would be predicted by the increased force required to simply push a broad head through a target.
F=MA ISN’T NEARLY ENOUGH TO INFORM YOU ABOUT PENETRATION!!! How much force was applied by the string ON AVERAGE to the knock? You can easily determine acceleration by from velocity, draw length and brace height, then use total arrow mass, and bingo, you now know your average force applied to the knock by the string. Obviously peak force occurred at full draw. Guess what? It won’t match the average force of your force draw curve! Why? Your bow limbs had to drive themselves forward too! The force applied to the knock during firing is less than the measured force of along your draw length!
bishoparchery.com
. I’m sure they are why would they have a website if they weren’t?
That said, I’ve heard a podcast or two involving complaints about high FOC and high wind. An arrow shaft that is misaligned with the arrow’s flight path is going to rob a lot of penetration.
