Suppose I draw back a bow as in bow and arrow. As I release the bow the 1 oz arrow accelerates over a period of about 22 inches and picks up about 36# of kinetic energy. So I figure the arrow weighs about 1 1/2# roughly under acceleration. The static weight of the bow is about 1 1/2# but is not accelerating because my hand is holding it. How much does the bow weigh durring the shot sequence??
My intuition tells me that the weight of the bow would not be relevant and the force against my hand would be equal and opposite to the forces of the arrow. But I have no way to figure it.
No. The arrow weighs 1oz during acceleration. Why would you assume it weighed more? And your units are all jacked up. A # or lb is a unit of force not a unit of energy.
I am talking about the influence it is having on the string and bow limbs which are the accelerating force. How would it pick up 36# ft pounds of kinetic energy if it only carried 1 oz?
Whoops I missed the edit window to actually answer the question. The bow weighs what it did. 1 1/2 #.
The kinetic energy is 1/2 mass * velocity^2. So the kinetic energy is coming from the speed as well as the mass. The influence it has on the bow would be related to conservation of momentum.
Some parts of the bow are undoubtedly accelerating. How else could they accelerate the arrow?
I can’t really parse your question into anything that would give a sensible answer. There are several problems in your statement, to wit:
[ul]
[li]“As I release the bow the 1 oz arrow accelerates over a period of about 22 inches”: A period is an interval of time; inches are a measurement of distance. The actual acceleration of the arrow is a function of the mass of the arrow and the rate of release of strain energy in the bow, and as an averaged value is therefore best described as a function of time. [/li][li]“and picks up about 36# of kinetic energy.”: Kinetic energy is in fundamental units of distance x mass, e.g. lbf･ft or N･m, which are often stated in joules (J) or horsepower (hp).[/li][li]“So I figure the arrow weighs about 1 1/2# roughly under acceleration.”: The mass (and therefore the weight) of the arrow is invariant. The bow is transferring momentum to the arrow (and a comparable amount in the opposite direction to the person using the bow) from the stored strain energy being in the flexure of the upper and low limb being released.[/li][li]“The static weight of the bow is about 1 1/2# but is not accelerating because my hand is holding it.”: The mass of the bow is invariant. The arrow is receiving momentum from the abovementioned release of stored strain energy in the bow. Elements of the bow are under acceleration even if the bow itself does not change overall position as measured at the grip. [/li][li]“How much does the bow weigh durring the shot sequence??”: The mass of the bow is invariant. The bow does convert the above mentioned stored strain energy into the momentum imparted upon the arrow and user. [/li][/ul]
Stranger
I didn’t pose the question properly. How much force is the bow applying to my bow hand opposite the accelerating arrow?
The bow would accelerate to about 185 fps over a period of about 22".
I am looking for how much force my bow hand would be experiencing durring the shot sequence.
Well, Newton sez momentum is equal in both directions. So, arrow mass * arrow velocity = bow mass * bow velocity (velocity measured in opposite directions). And bow force (on your hand) = bow mass / 2 * (bow velocity)^2. Do the math!
[ul]
[li]“The bow would accelerate to about 185 fps”: The average speed of the bow is approximately 0 unless the shooter is moving. The string is moving with the arrow until the nock is released, so you can assume that the peak speed speed of the string is approximately the same as that of the arrow. The upper and lower arms are moving forward at approximately the same speed as the string, and upward or downward at some additional speed.[/li][li]“over a period of about 22.”": Distance is not a “period” of time. [/li][li]“I am looking for how much force my bow hand would be experiencing durring the shot sequence.”: I assume you mean the peak dynamic force after release, which will occur at or (for compound bows) shortly after the nock is released. We cannot determine that from the information provided.[/ul] [/li]
If we assumed that the arrow was accelerated by the bow at a constant rate we could estimate the total impulse imparted into the arrow over the distance the bow acts upon it, and from there calculate the D’Alembert force (i.e. F = m･a). However, the answer you get won’t be sensible because the actual force sensed by the bowman will be dominated by the pre-release static force and the post-release dynamic flexure of the bow limbs. In fact, if you actually fire a bow, you will feel it try to pull forward after release because the stored energy of the bow that doesn’t go into the arrow is expended by the limbs reaching full forward extent, just as a spring when compressed and then released will jump forward from its own imparted momentum.
Stranger