JFK Assassination Forum
JFK Assassination Plus General Discussion & Debate => JFK Assassination Plus General Discussion And Debate => Topic started by: Charles Collins on March 05, 2025, 02:51:27 PM
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I have created a report for another Single Bullet Theory. I figured one SBT wasn't enough. ;)
You can find the report at this link:
https://sites.google.com/view/theothersbt/home
Please review this report if you are interested and let me know what you think. Thanks!
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I have created a report for another Single Bullet Theory. I figured one SBT wasn't enough. ;)
You can find the report at this link:
https://sites.google.com/view/theothersbt/home
Please review this report if you are interested and let me know what you think. Thanks!
You have put a good effort into your paper, Charles. While your theory that a wildly missed shot is better explained by accidental firing, the suggestion that Oswald would have been so careless as to fire accidentally seems just as highly improbable as an aimed shot that missed the car.
I do have a few questions, though.
1. You seem to equate velocity with energy as in: "The impact of the first ricochet (off of Elm Street) would have reduced the velocity considerably. We have to assume some values that seem reasonable. If we assume it lost about 31% of its velocity/energy,". Energy varies as the square of velocity so if the bullet loses 31% of its speed (retaining 69%) it ends up with .69^2=48% of its original energy, not 69%.
2. You say "the end point of the calculated trajectory as the bullet mark on the manhole cover, the graph shows an elevation drop of about 13.7-feet in about 83.3-yards (similar to what actually exists in Dealey Plaza). One of the main points that needs to be understood about the graphs is that the trajectories are essentially horizontal (level) for the first half of the distance, and slowly curve downward over the course of the second half.".
A fragment is still a projectile and follows the laws of physics. Neglecting air forces, any object drops at g=9.8 m/s^2 (32 f/s^2). So if it starts with zero vertical velocity, the distance, h, that it falls is proportional to time^2: h=gt^2/2. So t=√2h/√g. And since horizontal speed does not decrease much, time is proportional to distance travelled (t=d/v).
If the bullet hit the pavement and where you suggest and a fragment deflected at a 3 degree angle above the road surface so it started perfectly horizontal, in order to drop 13.7 feet (to the level of the manhole cover 83.3 yards or 250 feet) away, 2h/g=2x13.7/32=.856. So t=√.856=.925 seconds. So it would take .925 seconds to travel 250 feet. This means the fragment was travelling at an average speed of 250/.925 = 270 fps. So that fragment has lost 1-270/2165 = 87.5% of its initial speed, and 98% of its initial energy. If the bullet lost 98% of its energy in hitting the asphalt - ie. 600 [edit: this is wrong. 2000 fps is 600 m/s which works works out to 1800 J.] joules of energy would be deposited in a tiny area of the asphalt - there would have to be a significant movement of asphalt around that area.
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You have put a good effort into your paper, Charles. While your theory that a wildly missed shot is better explained by accidental firing, the suggestion that Oswald would have been so careless as to fire accidentally seems just as highly improbable as an aimed shot that missed the car.
I do have a few questions, though.
1. You seem to equate velocity with energy as in: "The impact of the first ricochet (off of Elm Street) would have reduced the velocity considerably. We have to assume some values that seem reasonable. If we assume it lost about 31% of its velocity/energy,". Energy varies as the square of velocity so if the bullet loses 31% of its speed (retaining 69%) it ends up with .69^2=48% of its original energy, not 69%.
2. You say "the end point of the calculated trajectory as the bullet mark on the manhole cover, the graph shows an elevation drop of about 13.7-feet in about 83.3-yards (similar to what actually exists in Dealey Plaza). One of the main points that needs to be understood about the graphs is that the trajectories are essentially horizontal (level) for the first half of the distance, and slowly curve downward over the course of the second half.".
A fragment is still a projectile and follows the laws of physics. Neglecting air forces, any object drops at g=9.8 m/s^2 (32 f/s^2). So if it starts with zero vertical velocity, the distance, h, that it falls is proportional to time^2: h=gt^2/2. So t=√2h/√g. And since horizontal speed does not decrease much, time is proportional to distance travelled (t=d/v).
If the bullet hit the pavement and where you suggest and a fragment deflected at a 3 degree angle above the road surface so it started perfectly horizontal, in order to drop 13.7 feet (to the level of the manhole cover 83.3 yards or 250 feet) away, 2h/g=2x13.7/32=.856. So t=√.856=.925 seconds. So it would take .925 seconds to travel 250 feet. This means the fragment was travelling at an average speed of 250/.925 = 270 fps. So that fragment has lost 1-270/2165 = 87.5% of its initial speed, and 98% of its initial energy. If the bullet lost 98% of its energy in hitting the asphalt - ie. 600 joules of energy would be deposited in a tiny area of the asphalt - there would have to be a significant movement of asphalt around that area.
Thanks for the reply Andrew.
You have put a good effort into your paper, Charles. While your theory that a wildly missed shot is better explained by accidental firing, the suggestion that Oswald would have been so careless as to fire accidentally seems just as highly improbable as an aimed shot that missed the car.
An accidental discharge can happen to anyone at any time. Even to highly trained, safety conscious professionals. We acquired a .45 caliber hole in our dining room table, chair, and floor when a gunsmith and safety instructor that we had known over 50-years had an accidental discharge. My idea that an accidental discharge happened to LHO could have been if the barrel of the rifle hit the box on the window sill. He most likely didn’t have an opportunity to actually practice his shots while in the sniper’s nest. And he most likely was looking through the scope as he began to try to track the moving target. If his finger was on the trigger the sudden unexpected collision with the box could cause the accidental discharge.
1. You seem to equate velocity with energy as in: "The impact of the first ricochet (off of Elm Street) would have reduced the velocity considerably. We have to assume some values that seem reasonable. If we assume it lost about 31% of its velocity/energy,". Energy varies as the square of velocity so if the bullet loses 31% of its speed (retaining 69%) it ends up with .69^2=48% of its original energy, not 69%.
Thanks I will revise the velocity/energy wording.
A fragment is still a projectile and follows the laws of physics. Neglecting air forces, any object drops at g=9.8 m/s^2 (32 f/s^2). So if it starts with zero vertical velocity, the distance, h, that it falls is proportional to time^2: h=gt^2/2. So t=√2h/√g. And since horizontal speed does not decrease much, time is proportional to distance travelled (t=d/v).
If the bullet hit the pavement and where you suggest and a fragment deflected at a 3 degree angle above the road surface so it started perfectly horizontal, in order to drop 13.7 feet (to the level of the manhole cover 83.3 yards or 250 feet) away, 2h/g=2x13.7/32=.856. So t=√.856=.925 seconds. So it would take .925 seconds to travel 250 feet. This means the fragment was travelling at an average speed of 250/.925 = 270 fps. So that fragment has lost 1-270/2165 = 87.5% of its initial speed, and 98% of its initial energy. If the bullet lost 98% of its energy in hitting the asphalt - ie. 600 joules of energy would be deposited in a tiny area of the asphalt - there would have to be a significant movement of asphalt around that area.
I said essentially level not level. I show the bullet fragment leaving Elm Street at a -2-degrees angle. This is one-degree above the road surface which slopes down at a -3-degrees angle. Also I have the fragment weighing 120-grains instead of the 161-grains it had as a whole bullet. The ballistic coefficient is a measure of efficiency the projectile has while in flight. The drastically lowered BC causes air friction to become a much larger factor in slowing the velocity over time and distance. I have circled the velocities and energy of the fragment beginning at the Elm Street mark and ending at the manhole cover for you. I believe that a significant amount of energy would have been lost due the fragmentation of the bullet when it hit Elm Street. And the amount of asphalt moved would depend on its hardness. Elm Street asphalt is relatively hard I do believe.
(https://i.vgy.me/VfqdZB.jpg)
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I believe that a significant amount of energy would have been lost due the fragmentation of the bullet when it hit Elm Street. And the amount of asphalt moved would depend on its hardness. Elm Street asphalt is relatively hard I do believe.
The hardness of the street determines whether, or how deep, the bullet penetrates. Asphalt, being a composite material, contains hard and soft particles. So the best way to determine what happens to asphalt struck by a 6.5 mm jacketed 10 gram bullet at 600 m/s striking asphalt at a 30 degree angle above horizontal would be experiment.
If it loses 98% of its energy in hitting the asphalt (which would be the case if the bullet fragments deflected from the asphalt were travelling at 270 fps, which is the case if a fragment dropped far enough to hit the manhole 250 distant) then it deposits all 98% of its incident energy in colliding with the asphalt: 1800 Joules. 1 gram of TNT releases 4184 Joules of energy when it explodes. So this would be like about half a gram of TNT being detonated in the asphalt surface. [Note in my earlier post I said 600 Joules - I was thinking of speed not energy - since corrected]
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The hardness of the street determines whether, or how deep, the bullet penetrates. Asphalt, being a composite material, contains hard and soft particles. So the best way to determine what happens to asphalt struck by a 6.5 mm jacketed 10 gram bullet at 600 m/s striking asphalt at a 30 degree angle above horizontal would be experiment.
If it loses 98% of its energy in hitting the asphalt (which would be the case if the bullet fragments deflected from the asphalt were travelling at 270 fps, which is the case if a fragment dropped far enough to hit the manhole 250 distant) then it deposits all 98% of its incident energy in colliding with the asphalt: 1800 Joules. 1 gram of TNT releases 4184 Joules of energy when it explodes. So this would be like about half a gram of TNT being detonated in the asphalt surface. [Note in my earlier post I said 600 Joules - I was thinking of speed not energy - since corrected]
Are you trying to calculate this in a vacuum? Or do you try to account for air resistance (drag)?
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The hardness of the street determines whether, or how deep, the bullet penetrates. Asphalt, being a composite material, contains hard and soft particles. So the best way to determine what happens to asphalt struck by a 6.5 mm jacketed 10 gram bullet at 600 m/s striking asphalt at a 30 degree angle above horizontal would be experiment.
If it loses 98% of its energy in hitting the asphalt (which would be the case if the bullet fragments deflected from the asphalt were travelling at 270 fps, which is the case if a fragment dropped far enough to hit the manhole 250 distant) then it deposits all 98% of its incident energy in colliding with the asphalt: 1800 Joules. 1 gram of TNT releases 4184 Joules of energy when it explodes. So this would be like about half a gram of TNT being detonated in the asphalt surface. [Note in my earlier post I said 600 Joules - I was thinking of speed not energy - since corrected]
Robert Frazier testified that the energy of the Carcano bullet as it leaves the muzzle is calculated to be about 1676 foot-pounds on average. This online calculator appears to agree with Frazier’s figure. It also agrees with the shooterscalculator.com figure for a 120-grain projectile at 1500 fps.
(https://i.vgy.me/vx4Xny.jpg)
This indicates to me that that particular fragment still contains about 36% of the initial energy of the bullet. There are obviously other fragments that make up the other 41-grains of that bullet that also still contain some of the initial energy. A significant amount of energy is required to fragment a 161-grain full metal jacketed bullet. So how much energy is transferred to the asphalt surface depends on the angle, the hardnesses of the two colliding bodies, etc.
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Are you trying to calculate this in a vacuum? Or do you try to account for air resistance (drag)?
Air resistance is not a significant factor in determining how much energy was lost in striking the asphalt if you want it to drop to the manhole and then to Main St. near Tague.
The air does not affect how long it takes to drop. So to drop 13.7 feet to strike the manhole will take just under a second. So the average speed has to be 270 fps. Now that could be because it started out faster than 270 fps and was movingly slower when it hit the manhole.
But you also want it to go another 141 feet and drop another 6.8 ft to the level of Main St. That takes .65 seconds so its average speed is over that distance is 141/.65=216 fps.
So before it hit the manhole it was going at least 216 fps and was travelling an average of 270 fps from where it hit the road. Assuming roughly constant resistance force vavg= (vf+ vi)/2. So if vf=216 and vavg=270 fps vi=324 fps. That is the maximum speed after hitting the road.
That means it retained only 324/2165=15% of it initial speed and .152=2.25% of its energy. So that means it lost 97.75% of its muzzle energy in hitting the road.
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Air resistance is not a significant factor in determining how much energy was lost in striking the asphalt if you want it to drop to the manhole and then to Main St. near Tague.
The air does not affect how long it takes to drop. So to drop 13.7 feet to strike the manhole will take just under a second. So the average speed has to be 270 fps. Now that could be because it started out faster than 270 fps and was movingly slower when it hit the manhole.
But you also want it to go another 141 feet and drop another 6.8 ft to the level of Main St. That takes .65 seconds so its average speed is over that distance is 141/.65=216 fps.
So before it hit the manhole it was going at least 216 fps and was travelling an average of 270 fps from where it hit the road. Assuming roughly constant resistance force vavg= (vf+ vi)/2. So if vf=216 and vavg=270 fps vi=324 fps. That is the maximum speed after hitting the road.
That means it retained only 324/2165=15% of it initial speed and .152=2.25% of its energy. So that means it lost 97.75% of its muzzle energy in hitting the road.
Andrew, you cannot ignore the air resistance. Here’s a little blurb that might surprise you.
“The invention of the ballistic pendulum by the English ballistician Benjamin Robins in 1740 had led to the astounding discovery (at that time) that the drag force on a bullet was many times more powerful than the force due to gravity, and that it changed markedly with bullet velocity.”
https://www.sierrabullets.com/exterior-ballistics/2-0-the-ballistic-coefficient/ (https://www.sierrabullets.com/exterior-ballistics/2-0-the-ballistic-coefficient/)
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Here’s an image showing the drag coefficients of different shaped objects. A bullet is designed to be streamlined to a certain extent. You can see for yourself by comparing the drag coefficients of the different shaped objects.
(https://i.vgy.me/jgDWyO.jpg)
An irregular shaped fragment, that is also tumbling, from the impact with Elm Street would have an extremely high drag coefficient due to turbulence. The drag would be magnified as the fragment transitioned from supersonic to subsonic velocity due to naturally increased turbulence in the trans-sonic velocity range. The net effect would be similar to deploying a relatively large parachute (think fighter jet airplane landing then slowing down with the assistance of parachutes).
A large drag coefficient means a low ballistic coefficient. I used a very low ballistic coefficient of 0.00405 in the online ballistic calculator. There is no way we can know exactly what the values actually were in Dealey Plaza. I indicated that we have to assume some reasonable values. I came up with what seems reasonable values to me, just to show that this idea is feasible.
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Andrew, you cannot ignore the air resistance. Here’s a little blurb that might surprise you.
“The invention of the ballistic pendulum by the English ballistician Benjamin Robins in 1740 had led to the astounding discovery (at that time) that the drag force on a bullet was many times more powerful than the force due to gravity, and that it changed markedly with bullet velocity.”
No surprise at all. Drag is huge. FBI expert Robert Frazier used 1 fps per foot so over 165 feet of travel bullet speed would go from 2165 fps to 2000 fps. That is a deceleration of roughly 1fps/(1/2000s) = 2000 fps2. Gravity is only 32 fps2.
The drag is a function of v2 where v is the velocity in the direction of motion through the air. That does not affect the time a bullet takes to drop a short distance due to gravity because it's falling speed is so much slower. That's all I am saying.
In the time it takes the bullet to travel 200 feet (.1 sec at average 2000 fps)
the bullet will be falling at 3.2 fps and will have dropped .16 ft. So in determining falling time you can ignore air resistance.
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No surprise at all. Drag is huge. FBI expert Robert Frazier used 1 fps per foot so over 165 feet of travel bullet speed would go from 2165 fps to 2000 fps. That is a deceleration of roughly 1fps/(1/2000s) = 2000 fps2. Gravity is only 32 fps2.
The drag is a function of v2 where v is the velocity in the direction of motion through the air. That does not affect the time a bullet takes to drop a short distance due to gravity because it's falling speed is so much slower. That's all I am saying.
In the time it takes the bullet to travel 200 feet (.1 sec at average 2000 fps)
the bullet will be falling at 3.2 fps and will have dropped .16 ft. So in determining falling time you can ignore air resistance.
No, that is not all you were trying to say. You were trying to calculate the energy transferred to the asphalt without taking into consideration the energy lost due to drag.
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No, that is not all you were trying to say. You were trying to calculate the energy transferred to the asphalt without taking into consideration the energy lost due to drag.
No. I am just pointing out that if you want the bullet fragment, after hitting the asphalt, to fall 13.7 feet vertically over a horizontal range of 250 feet, (and then fall another 6.8 feet over the next 141 ft) it can’t be travelling much more than 300 fps after hitting the asphalt.
I took into account the loss of energy due to air resistance. The muzzle speed was 2165 fps or 660 m/s so the 10g.bullet initially had mv^2/2 energy=(.005(660)^2)=2,178 J. Just before hitting the road it had slowed to 2000 fps due to air resistance. So its energy on impact with the road was .005(610)^2=1,860 J. Upon hitting the road, the bullet fragmented but all fragments kept moving slowly enough to fall 13.7 feet over a range of 250 feet which means it had an average speed of 270 fps.
Even if it was moving at 440 fps (134 m/s) after hitting the road and slowed to 100 fps just before hitting the manhole (but then it could not have hit the road near Tague 141 ft away) that means it had retained only (134/610)^2=0.05=5% of its original energy. So it deposited 95% of its 1860 Joules of energy (1767 J) in colliding with the road. I am saying it was a bit more than that - about 1800 J.
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No. I am just pointing out that if you want the bullet fragment, after hitting the asphalt, to fall 13.7 feet vertically over a horizontal range of 250 feet, (and then fall another 6.8 feet over the next 141 ft) it can’t be travelling much more than 300 fps after hitting the asphalt.
I took into account the loss of energy due to air resistance. The muzzle speed was 2165 fps or 660 m/s so the 10g.bullet initially had mv^2/2 energy=(.005(660)^2)=2,178 J. Just before hitting the road it had slowed to 2000 fps due to air resistance. So its energy on impact with the road was .005(610)^2=1,860 J. Upon hitting the road, the bullet fragmented but all fragments kept moving slowly enough to fall 13.7 feet over a range of 250 feet which means it had an average speed of 270 fps.
Even if it was moving at 440 fps (134 m/s) after hitting the road and slowed to 100 fps just before hitting the manhole (but then it could not have hit the road near Tague 141 ft away) that means it had retained only (134/610)^2=0.05=5% of its original energy. So it deposited 95% of its 1860 Joules of energy (1767 J) in colliding with the road. I am saying it was a bit more than that - about 1800 J.
Andrew, you are trying to over-simplify more than one aspect. And you are trying to compare apples to oranges. Your methods do not work for this application. I understand the physics behind firing a bullet level and dropping a bullet straight down and they both hit a level surface at the same time. You are trying to specify a slower velocity for the fragment to give it enough time to drop 13.7-feet. However that is simply not the situation we are faced with in this case.
First of all, as I have already explained, the fragment leaves the surface of Elm Street at a negative 2-degrees angle (not level). Therefore, at a distance of 250-feet it is going to be at an elevation that is 8.7-feet lower than it was at the Elm Street mark without any help from gravity due to simple geometry. Also, unlike the dropped bullet, due to the downward component of the direction (-2-degrees) of the fragment's velocity when it leaves the Elm Street mark the fragment is not starting from zero vertical velocity.
So the fragment only needs to drop approximately 5-feet due to gravity (not 13.7-feet). Plus it has a "running start" compared to a dropped bullet. You can compute all of this if you wish. However, I will likely continue to believe the ballistic calculators.
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Andrew, you are trying to over-simplify more than one aspect. And you are trying to compare apples to oranges. Your methods do not work for this application. I understand the physics behind firing a bullet level and dropping a bullet straight down and they both hit a level surface at the same time. You are trying to specify a slower velocity for the fragment to give it enough time to drop 13.7-feet. However that is simply not the situation we are faced with in this case.
First of all, as I have already explained, the fragment leaves the surface of Elm Street at a negative 2-degrees angle (not level). Therefore, at a distance of 250-feet it is going to be at an elevation that is 8.7-feet lower than it was at the Elm Street mark without any help from gravity due to simple geometry. Also, unlike the dropped bullet, due to the downward component of the direction (-2-degrees) of the fragment's velocity when it leaves the Elm Street mark the fragment is not starting from zero vertical velocity.
So the fragment only needs to drop approximately 5-feet due to gravity (not 13.7-feet). Plus it has a "running start" compared to a dropped bullet. You can compute all of this if you wish. However, I will likely continue to believe the ballistic calculators.
Where does your paper mention a 2 degree downward angle? Your charts indicate a slight upward angle.
In any event it doesn’t matter much. A fall of 5 feet requires .56 seconds so the average speed will be 250/.56=447 fps. And let’s say it leaves the asphalt at 600 fps and when it strikes the manhole it is down to around 300 fps. At 600 fps leaving the asphalt it has (600/2000)^2=0.09=9% of the energy it had before hitting the asphalt. So it lost 91% of its 1860 J=1693J. in striking the asphalt. That still results in a lot of energy transferred to the asphalt.
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Where does your paper mention a 2 degree downward angle? Your charts indicate a slight upward angle.
In any event it doesn’t matter much. A fall of 5 feet requires .56 seconds so the average speed will be 250/.56=447 fps. And let’s say it leaves the asphalt at 600 fps and when it strikes the manhole it is down to around 300 fps. At 600 fps leaving the asphalt it has (600/2000)^2=0.09=9% of the energy it had before hitting the asphalt. So it lost 91% of its 1860 J=1693J. in striking the asphalt. That still results in a lot of energy transferred to the asphalt.
Where does your paper mention a 2 degree downward angle? Your charts indicate a slight upward angle.
The below image is from the report. I posted it here in my first reply to your response. I pointed out the -2-degrees angle in my reply. I drew a red arrow on the graph to make it easy to find.
(https://i.vgy.me/VfqdZB.jpg)
These graphs are designed to aid in sighting in guns and their sights, and for calculating hold-overs. They are not really made for ricochets. This is an improvised attempt at using one of the graphs for the purpose of demonstrating the idea of the other single bullet theory. I agree that if you are only looking at the graph itself it appears level, but that is not the angle that was input. This is apparently a limitation of the graph. Also, I may need to adjust for only a 5-foot drop, the jury is still out on that. I appreciate your input. I am still experimenting with actual results of actual ricochets.
You are still ignoring the fact that the vertical component of the direction (-2-degrees) gives the ricocheted fragment a “running start” compared to dropping a bullet from the same height.
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Where does your paper mention a 2 degree downward angle? Your charts indicate a slight upward angle.
The below image is from the report. I posted it here in my first reply to your response. I pointed out the -2-degrees angle in my reply. I drew a red arrow on the graph to make it easy to find.
(https://i.vgy.me/VfqdZB.jpg)
These graphs are designed to aid in sighting in guns and their sights, and for calculating hold-overs. They are not really made for ricochets. This is an improvised attempt at using one of the graphs for the purpose of demonstrating the idea of the other single bullet theory. I agree that if you are only looking at the graph itself it appears level, but that is not the angle that was input. This is apparently a limitation of the graph. Also, I may need to adjust for only a 5-foot drop, the jury is still out on that. I appreciate your input. I am still experimenting with actual results of actual ricochets.
You are still ignoring the fact that the vertical component of the direction (-2-degrees) gives the ricocheted fragment a “running start” compared to dropping a bullet from the same height.
I am not ignoring that at all. That just gives it the constant rate of drop of v0sin(2) resulting in a drop of 8.7 feet over 250 feet. The additional drop of 5 feet requires t=sqrt(2h/g)=(2x5/32)^(1/2)=0.56 seconds. That is what limits the speed of the fragment.
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I am not ignoring that at all. That just gives it the constant rate of drop of v0sin(2) resulting in a drop of 8.7 feet over 250 feet. The additional drop of 5 feet requires t=sqrt(2h/g)=(2x5/32)^(1/2)=0.56 seconds. That is what limits the speed of the fragment.
Sorry, but I disagree. The dropped bullet accelerates from zero velocity without any other significant force other than gravity. The ricochet has a "running start" due to its momentum. If its velocity is 1500-fps, then I believe that 2/90 = .0222222 X 1500 = 33.33 fps vertically is its "running start."
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Sorry, but I disagree. The dropped bullet accelerates from zero velocity without any other significant force other than gravity. The ricochet has a "running start" due to its momentum. If its velocity is 1500-fps, then I believe that 2/90 = .0222222 X 1500 = 33.33 fps vertically is its "running start."
A projectile follows the following path defined by its horizontal (x) and vertical coordinates (y):
x=x0 + v0cos(θ)t
y= y0 + v0sin(θ)t - (1/2)gt2
See: http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#tra5 (http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#tra5)
If you apply that to your situation,
x=x0=0;
(1) v0cos(2)t=250 ft.
y=y0=0;
v0sin(2)t=250 sin(2)=8.7 ft
(2) (1/2)gt2=16t2=5 ft
This last equation (2) means t= √(5/16)=0.56 seconds
Substituting into (1) gives v0=250/(.56x.9994)=447 fps
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A projectile follows the following path defined by its horizontal (x) and vertical coordinates (y):
x=x0 + v0cos(θ)t
y= y0 + v0sin(θ)t - (1/2)gt2
See: http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#tra5 (http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#tra5)
If you apply that to your situation,
x=x0=0;
(1) v0cos(2)t=250 ft.
y=y0=0;
v0sin(2)t=250 sin(2)=8.7 ft
(2) (1/2)gt2=16t2=5 ft
This last equation (2) means t= √(5/16)=0.56 seconds
Substituting into (1) gives v0=250/(.56x.9994)=447 fps
That’s a very interesting and useful link you posted Andrew. Thank you. At first glance, it appears that the different calculators are intended for level or upwards directions. However, I tried a -2-degree launch angle in this one and a shorter time interval than you keep trying to use and got some results that appear to be “in the ballpark” of what is needed for the situation in Dealey Plaza. And hopefully, the numerical results in this image illustrate what I have been trying to say.
(https://i.vgy.me/RA8rOV.jpg)
Although I did indicate the ballistic graphs in my report were not intended to be totally accurate (only to show the concepts), I think the difference between what you are calculating and what I am coming up with is larger than it might should be. So, I think that one of us is doing something wrong. I want the report to be honest and reasonably accurate. So your input is valuable to me and appreciated. I plan to double check the elevations I used and look for any errors on my part. Please let me know what you think about the above calculator results. Thanks again.
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A projectile follows the following path defined by its horizontal (x) and vertical coordinates (y):
x=x0 + v0cos(θ)t
y= y0 + v0sin(θ)t - (1/2)gt2
See: http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#tra5 (http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#tra5)
If you apply that to your situation,
x=x0=0;
(1) v0cos(2)t=250 ft.
y=y0=0;
v0sin(2)t=250 sin(2)=8.7 ft
(2) (1/2)gt2=16t2=5 ft
This last equation (2) means t= √(5/16)=0.56 seconds
Substituting into (1) gives v0=250/(.56x.9994)=447 fps
I did find that in reviewing the elevations in Dealey Plaza and using interpolation between two measured points’ elevations, instead of just the closest point, the difference in elevation between the bullet strike and the manhole cover is actually closer to 12.7-feet instead of the 13.7-feet originally indicated in my report.
I also want to point out that the equations and calculators on the webpage that you linked to do not include anything for air resistance. You still have not addressed the air resistance that the tumbling fragment would have. You did address the air resistance of the bullet before it struck Elm Street, but nothing at all afterwards. Here’s a blurb from the webpage you linked to and that contains the equations.
“ The motion of an object under the influence of gravity is determined completely by the acceleration of gravity, its launch speed, and launch angle provided air friction is negligible.”
I think that the air resistance of a tumbling fragment is significant, not negligible. Even you acknowledged that the air resistance is “huge”. We need to take the air resistance into consideration. I think that the ballistic calculator that I used in my report allows this. I used a very low ballistic coefficient input into that calculator to account for high air resistance. That is where I believe the difference in our results lies.
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I did find that in reviewing the elevations in Dealey Plaza and using interpolation between two measured points’ elevations, instead of just the closest point, the difference in elevation between the bullet strike and the manhole cover is actually closer to 12.7-feet instead of the 13.7-feet originally indicated in my report.
I also want to point out that the equations and calculators on the webpage that you linked to do not include anything for air resistance. You still have not addressed the air resistance that the tumbling fragment would have. You did address the air resistance of the bullet before it struck Elm Street, but nothing at all afterwards. Here’s a blurb from the webpage you linked to and that contains the equations.
“ The motion of an object under the influence of gravity is determined completely by the acceleration of gravity, its launch speed, and launch angle provided air friction is negligible.”
I think that the air resistance of a tumbling fragment is significant, not negligible. Even you acknowledged that the air resistance is “huge”. We need to take the air resistance into consideration. I think that the ballistic calculator that I used in my report allows this. I used a very low ballistic coefficient input into that calculator to account for high air resistance. That is where I believe the difference in our results lies.
Air resistance in the direction of travel is huge. So air resistance to horizontal movement is significant especially at high speeds around 2000 fps. Gravity is not in the direction of travel. There is no difference, as you acknowledge, in time to drop a given distance for a bullet fired from a gun and a bullet that is just dropped. Air resistance for the dropped bullet is negligible over a drop distance that we are talking about here. That is why air resistance does not affect the time required for a bullet to drop such a short distance.
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Air resistance in the direction of travel is huge. So air resistance to horizontal movement is significant especially at high speeds around 2000 fps. Gravity is not in the direction of travel. There is no difference, as you acknowledge, in time to drop a given distance for a bullet fired from a gun and a bullet that is just dropped. Air resistance for the dropped bullet is negligible over a drop distance that we are talking about here. That is why air resistance does not affect the time required for a bullet to drop such a short distance.
Gravity is not in the direction of travel.
What!? ??? Gravity is downward toward the center of the earth. The ricocheted fragment has a downward component right from the time it leaves the Elm Street surface. That downward component increases over time. Especially in the second half of the calculated trajectory (parabolic curve). So I think that you are just plain wrong.
Some information from “Modern Practical Ballistics” by Arthur J. Pejsa:
Back in the 1730s Benjamin Robbins invented the ballistic pendulum and showed that the drag force (air resistance) is fifty to one hundred times as great as the force of gravity. Additionally, at speeds approaching the speed of sound, there is a rapid and dramatic twofold to threefold increase in air drag; between approximately 1000 and 1200 fps, air drag is found to increase with about the fifth or sixth power of the air speed rather than with the square, or second power as it does at lower speeds. This is the infamous “sonic barrier” we hear of.
I am currently building and experimenting with a DIY ballistic pendulum. So far, ~30-degrees ricochets off of a concrete surface appear to retain the vast majority of their initial velocities. This agrees with the results of the high speed filmed 30-degrees ricochets that I included screenshots from in my report. You can search and find and watch that video on YouTube if you use the information at the bottom of the last screenshot. They show the various penetration depths of the various angle ricochets in ballistic gel.
I intend to continue to study and learn more from Pejsa’s book and my experiments. So far I have to disagree with your assessment and claims based on what I have learned so far. Again, I do appreciate your input it tends to give me incentive to learn more about these things.
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Gravity is not in the direction of travel.
What!? ??? Gravity is downward toward the center of the earth. The ricocheted fragment has a downward component right from the time it leaves the Elm Street surface. That downward component increases over time. Especially in the second half of the calculated trajectory (parabolic curve). So I think that you are just plain wrong.
Some information from “Modern Practical Ballistics” by Arthur J. Pejsa:
Back in the 1730s Benjamin Robbins invented the ballistic pendulum and showed that the drag force (air resistance) is fifty to one hundred times as great as the force of gravity. Additionally, at speeds approaching the speed of sound, there is a rapid and dramatic twofold to threefold increase in air drag; between approximately 1000 and 1200 fps, air drag is found to increase with about the fifth or sixth power of the air speed rather than with the square, or second power as it does at lower speeds. This is the infamous “sonic barrier” we hear of.
I am currently building and experimenting with a DIY ballistic pendulum. So far, ~30-degrees ricochets off of a concrete surface appear to retain the vast majority of their initial velocities. This agrees with the results of the high speed filmed 30-degrees ricochets that I included screenshots from in my report. You can search and find and watch that video on YouTube if you use the information at the bottom of the last screenshot. They show the various penetration depths of the various angle ricochets in ballistic gel.
I intend to continue to study and learn more from Pejsa’s book and my experiments. So far I have to disagree with your assessment and claims based on what I have learned so far. Again, I do appreciate your input it tends to give me incentive to learn more about these things.
At a 2 degree downward slope, the downward component initially is -.035 times the horizontal component. So even at 2000 fps the downward speed would be no more than -70 fps. And after .1 second it would be -73.2 fps. (vy=v0y-gt)
If there is more drag it will take longer to fall a given distance so it would have to be moving horizontally at an even slower speed for it to fall 5 feet (or 4 feet now with your revision) by taking into account vertical drag due to air resistance.
Projectile motion is fairly basic if one does not take into account air resistance. That’s why it is a first year physics subject. Fluid dynamics is typically a third or fourth year subject as it is very complicated.
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At a 2 degree downward slope, the downward component initially is -.035 times the horizontal component. So even at 2000 fps the downward speed would be no more than -70 fps. And after .1 second it would be -73.2 fps. (vy=v0y-gt)
If there is more drag it will take longer to fall a given distance so it would have to be moving horizontally at an even slower speed for it to fall 5 feet (or 4 feet now with your revision) by taking into account vertical drag due to air resistance.
Projectile motion is fairly basic if one does not take into account air resistance. That’s why it is a first year physics subject. Fluid dynamics is typically a third or fourth year subject as it is very complicated.
A total velocity of 1500 fps with a downward component of 0.035 X 1500 = 52.5 fps. In 0.2419 seconds it would drop approximately the required 12.7-feet. The vertical component of the drag force would be acting to slow this descent while gravity would be acting to speed it up. Since the vertical component of the drag force would be stronger than gravity, the net effect would be deceleration of the descent. But, the point I have been making is that since there is a vertical velocity to begin with, the time required to accelerate from zero velocity (of a dropped bullet) is eliminated. Here is an image that shows the dropped bullet’s much smaller velocity and distance at the same 0.2419 seconds elapsed time. That is the running start I have been describing.
(https://i.vgy.me/eXQ6VI.jpg)
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Oswald most likely was looking through the scope as he began to try to track the moving target.
Why in the world would former sharpshooter Oswald have been looking through the crummy, dimly lit scope?
If he had "tracked" JFK's head through his scope as the limo was coming straight towards him on Houston Street, he not only would have been spotted by the Secret Service agents in the follow-up car, but would have shot at JFK's huge, looming, enticing-through-the-scope head.
However, if he *was* "tacking" JFK's ever-more enticing head through his scope while the limo was still on Houston Street and 1) the Secret Service agents *somehow* didn't notice him and 2) he *somehow* resisted the temptation squeeze off a round at JFK's large, looming, ever-enticing, impossible-to-miss head, when the limo started turning onto Elm Street, he may have tried to switch over to his much-more-practical-at-that-close-range iron sights and . . . gasp . . . . lost his bead on JFK's head in the process.
If he *didn't* switch over to his iron sights when the limo turned onto Elm Street but stupidly continued "tracking" JFK's head through his scope, that in-and-of-itself could explain how he managed to miss everything with his first shot.
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Why in the world would he have been looking through the scope?
If he had "tracked" JFK's head with his scope as the limo was coming towards him on Houston Street, he not only would have been spotted by the Secret Service agents in the follow-up car, but would have shot at JFK's huge, looming, enticing-through-the-scope head.
However, if he *was* "tacking" JFK's ever-more enticing head through his scope while the limo was still on Houston Street and 1) the Secret Service agents *somehow* didn't notice him and 2) he *somehow* resisted the temptation squeeze off a round at JFK's large, looming, ever-enticing, impossible-to-miss head, when the limo started turning onto Elm Street, he may have tried to switch over to his much-more-practical-at-that-close-range iron sights and . . . gasp . . . . lost his bead on JFK's head in the process.
If he *didn't* switch over to his iron sights when the limo turned onto Elm Street but stupidly continued "tracking" JFK's head through his scope, that in-and-of-itself could explain how he managed to miss everything with his first shot.
Why in the world would he have been looking through the scope?
A scope typically provides a faster and more accurate way to shoot.
If he had "tracked" JFK's head with his scope as the limo was coming towards him on Houston Street, he not only would have been spotted by the Secret Service agents in the follow-up car, but would have shot at JFK's huge, looming, enticing-through-the-scope head.
I said nothing about tracking while the limo was still on Houston Street. This all begins as the limo was turning onto Elm Street.
However, if he *was* "tacking" JFK's ever-more enticing head through his scope while the limo was still on Houston Street and 1) the Secret Service agents *somehow* didn't notice him and 2) he *somehow* resisted the temptation squeeze off a round at JFK's large, looming, ever-enticing, impossible-to-miss head, when the limo started turning onto Elm Street, he may have tried to switch over to his much-more-practical-at-that-close-range iron sights and . . . gasp . . . . lost his bead on JFK's head in the process.
First of all, a 4X power scope is not too powerful to use at a distance of about 100-feet. You can ask many dead squirrels about that. Also, it really doesn't matter whether he was using the scope or the fixed iron sights. The point is that he would have been trying to acquire the target in his sights and not paying attention to the box on the window sill.
If he *didn't* switch over to his iron sights when the limo turned onto Elm Street but stupidly continued "tracking" JFK's head through his scope, that in-and-of-itself could explain how he managed to miss everything with his first shot
No it doesn't. LHO apparently bought the scope as an extra cost item to use for a shot at General Walker. That shot at Walker was about the same distance (100-feet) as an early shot at the limo. LHO apparently spent time shooting the rifle/scope combo prior to the Walker event. I believe he would have zeroed the scope for the same distance he planned to shoot at. So it seems to me that using the scope makes the most sense.
Here's a clip that shows what I think may have happened. It starts with a plan to begin shooting as soon as the limo clears out from under the tree. It takes a second or two or three to raise the rifle, acquire the target in the sights, and pull the trigger. So I think that he would have begun that process while the limo was directly below him and once he acquired the target he would track it as best he could while it was passing under the tree (from his point of view). He probably didn't have a chance to practice this except in his mind. So, when first lowering the rifle towards the limo, if it suddenly and unexpectedly hit the box on the window sill (as depicted in the clip) and his finger was on the trigger, an inadvertent (accidental discharge) shot could have occurred. That is what I believe happened.
(https://i.vgy.me/iSyyAf.gif)
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Why in the world would he have been looking through the scope?
A scope typically provides a faster and more accurate way to shoot.
If he had "tracked" JFK's head with his scope as the limo was coming towards him on Houston Street, he not only would have been spotted by the Secret Service agents in the follow-up car, but would have shot at JFK's huge, looming, enticing-through-the-scope head.
I said nothing about tracking while the limo was still on Houston Street. This all begins as the limo was turning onto Elm Street.
However, if he *was* "tacking" JFK's ever-more enticing head through his scope while the limo was still on Houston Street and 1) the Secret Service agents *somehow* didn't notice him and 2) he *somehow* resisted the temptation squeeze off a round at JFK's large, looming, ever-enticing, impossible-to-miss head, when the limo started turning onto Elm Street, he may have tried to switch over to his much-more-practical-at-that-close-range iron sights and . . . gasp . . . . lost his bead on JFK's head in the process.
First of all, a 4X power scope is not too powerful to use at a distance of about 100-feet. You can ask many dead squirrels about that. Also, it really doesn't matter whether he was using the scope or the fixed iron sights. The point is that he would have been trying to acquire the target in his sights and not paying attention to the box on the window sill.
If he *didn't* switch over to his iron sights when the limo turned onto Elm Street but stupidly continued "tracking" JFK's head through his scope, that in-and-of-itself could explain how he managed to miss everything with his first shot
No it doesn't. LHO apparently bought the scope as an extra cost item to use for a shot at General Walker. That shot at Walker was about the same distance (100-feet) as an early shot at the limo. LHO apparently spent time shooting the rifle/scope combo prior to the Walker event. I believe he would have zeroed the scope for the same distance he planned to shoot at. So it seems to me that using the scope makes the most sense.
Here's a clip that shows what I think may have happened. It starts with a plan to begin shooting as soon as the limo clears out from under the tree. It takes a second or two or three to raise the rifle, acquire the target in the sights, and pull the trigger. So I think that he would have begun that process while the limo was directly below him and once he acquired the target he would track it as best he could while it was passing under the tree (from his point of view). He probably didn't have a chance to practice this except in his mind. So, when first lowering the rifle towards the limo, if it suddenly and unexpectedly hit the box on the window sill (as depicted in the clip) and his finger was on the trigger, an inadvertent (accidental discharge) shot could have occurred. That is what I believe happened.
(https://i.vgy.me/iSyyAf.gif)
Not wanting to have to wade through the whole thread to see if you've already answered this question, please tell me what your first shot equates to Zapruder-frame-wise.
Thanks.
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Not wanting to have to wade through the whole thread to see if you've already answered this question, please tell me what your first shot equates to Zapruder-frame-wise.
Thanks.
If you read the report you might remember this image showing a clear path for the bullet/fragment(s) at about Z133. This appears to be true for a period of time from shortly before to shortly after Z133.
(https://i.vgy.me/FYwbkG.jpg)
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If you read the report you might remember this image showing a clear path for the bullet/fragment(s) at about Z133. This appears to be true for a period of time from shortly before to shortly after Z133.
(https://i.vgy.me/FYwbkG.jpg)
No need to be snide, Chuck.
Why don't you like Roselle's and Scearce's hypothetical "Z-124"?
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No need to be snide, Chuck.
Why don't you like Roselle's and Scearce's hypothetical "Z-124"?
I wasn’t being snide. Whatever gave you that idea?
I didn’t say anything about the Roselle and Scearce idea. Why do you think I don’t like it?
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A total velocity of 1500 fps with a downward component of 0.035 X 1500 = 52.5 fps. In 0.2419 seconds it would drop approximately the required 12.7-feet. The vertical component of the drag force would be acting to slow this descent while gravity would be acting to speed it up. Since the vertical component of the drag force would be stronger than gravity, the net effect would be deceleration of the descent. But, the point I have been making is that since there is a vertical velocity to begin with, the time required to accelerate from zero velocity (of a dropped bullet) is eliminated. Here is an image that shows the dropped bullet’s much smaller velocity and distance at the same 0.2419 seconds elapsed time. That is the running start I have been describing.
(https://i.vgy.me/eXQ6VI.jpg)
The running start is what reduces the fall due to gravity from 13.7 feet to 5 feet (or 4). You can't count it twice. The rate of vertical descent increases that vertical descent speed (v0sin(2)) by gt where g=32 ft/sec2 and t=time in seconds. The time, t, is horizontal distance/horizontal speed.
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"If you read the report you might remember..."
I know.
Confusing answer.
Regardless, why don't you just answer my question?
You need to consider re-wording your question. It is a loaded question as it is.
But I can probably save you some time and effort by telling you that my report on “the other single bullet theory” is about where that bullet went. The Roselle/Scearce idea is about when the shot occurred. Two different subjects.
I have given you an answer in the form of a range of time when there was a clear path for the bullet to cross Elm Street. Any time within that range is okay with me. I do believe that the Roselle/Scearce time is within the range that I already indicated.
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The running start is what reduces the fall due to gravity from 13.7 feet to 5 feet (or 4). You can't count it twice. The rate of vertical descent increases that vertical descent speed (v0sin(2)) by gt where g=32 ft/sec2 and t=time in seconds. The time, t, is horizontal distance/horizontal speed.
I am not counting anything twice. If the fragment already has a vertical velocity, gravity adds to it.
Regardless, the difference in our thinking is your lack of applying the air resistance to the fragment after it ricochets off of Elm Street.
We already established that simple geometry and a 2-degree downward ricochet angle reduces the amount of drop necessary to approximately 47.64” (12.7’-8.73’). Your claim that the fragment has to be going at a drastically reduced speed to have the time necessary to drop that far does seem to be correct. However, your claim that the necessary velocity and energy loss would have to be due to the collision with the pavement on Elm Street is not correct. The reason it is not correct is that you have not taken into account the air resistance force on the fragment after it ricochets.
I believe that I have figured out a few things about the ballistic calculators. And I have found a different ballistic calculator that shows more information. Here are a couple of images that show how the air resistance affects things.
First, here is a table showing how quickly the velocity and energy values are reduced. The distance from the mark on Elm Street to the manhole cover we have been using using is 250’ (~83.33-yards). Based on that, and looking at the table, we can see that after a distance of only 16-yards the velocity is already reduced by about half of the initial 1500 fps. And the elapsed time is only about 0.05-seconds. Wow, that is due to the huge (your word) effect of the air resistance particularly during the trans-sonic stage. And this already solves the “mystery” of a lot of our differences.
(https://i.vgy.me/VGOwn5.png)
Next are some graphs. I have determined that the “level” line at the top of the top left graph represents the line of sight. I have set the height of the “rifle” sights at zero above the bore. Therefore that line represents sighting through the bore of a rifle aimed at 2-degrees below level (the same as the direction of the ricochet off of Elm Street). So the necessary drop is ~47.64”. The ballistic calculator indicates a drop of ~46.8”. Therefore we are within an inch and that should be close enough for the purposes of this discussion.
(https://i.vgy.me/ZU6uIw.png)
I hope this helps you understand the importance of accounting for the air resistance. And that the needed reduction in the velocity and energy of the fragment is not totally due to a loss at the time of the collision with the pavement on Elm Street. If this were taking place in a vacuum, yes; but not in the atmosphere in Dealey Plaza on 11/22/63.
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I am not counting anything twice. If the fragment already has a vertical velocity, gravity adds to it.
Yes. But the time it takes to add a distance h is ½gt2 so for an additional 4 feet, t=√(2x4/32)=.5 seconds. That is how long it takes for the fragment to travel from the place where it strikes the asphalt to where it strikes the manhole (in your scenario). So its average speed over that 250 feet is 250/.5=500 fps.
Regardless, the difference in our thinking is your lack of applying the air resistance to the fragment after it ricochets off of Elm Street.
We already established that simple geometry and a 2-degree downward ricochet angle reduces the amount of drop necessary to approximately 47.64” (12.7’-8.73’). Your claim that the fragment has to be going at a drastically reduced speed to have the time necessary to drop that far does seem to be correct. However, your claim that the necessary velocity and energy loss would have to be due to the collision with the pavement on Elm Street is not correct. The reason it is not correct is that you have not taken into account the air resistance force on the fragment after it ricochets.
I believe that I have figured out a few things about the ballistic calculators. And I have found a different ballistic calculator that shows more information. Here are a couple of images that show how the air resistance affects things.
First, here is a table showing how quickly the velocity and energy values are reduced. The distance from the mark on Elm Street to the manhole cover we have been using using is 250’ (~83.33-yards). Based on that, and looking at the table, we can see that after a distance of only 16-yards the velocity is already reduced by about half of the initial 1500 fps. And the elapsed time is only about 0.05-seconds. Wow, that is due to the huge (your word) effect of the air resistance particularly during the trans-sonic stage. And this already solves the “mystery” of a lot of our differences.
(https://i.vgy.me/VGOwn5.png)
Next are some graphs. I have determined that the “level” line at the top of the top left graph represents the line of sight. I have set the height of the “rifle” sights at zero above the bore. Therefore that line represents sighting through the bore of a rifle aimed at 2-degrees below level (the same as the direction of the ricochet off of Elm Street). So the necessary drop is ~47.64”. The ballistic calculator indicates a drop of ~46.8”. Therefore we are within an inch and that should be close enough for the purposes of this discussion.
(https://i.vgy.me/ZU6uIw.png)
I hope this helps you understand the importance of accounting for the air resistance. And that the needed reduction in the velocity and energy of the fragment is not totally due to a loss at the time of the collision with the pavement on Elm Street. If this were taking place in a vacuum, yes; but not in the atmosphere in Dealey Plaza on 11/22/63.
If the fragment:
- started out at 1500 fps, meaning the bullet would have lost only 44% of its incident energy of 1860 J (assuming the bullet struck asphalt at 2000 fps) - which still works out to 800 Joules - and
- averaged 500 fps over that 250 ft distance, and
- was able to drop a further 6.8 feet over the next 141 feet
what do you say was its speed over the last 141 ft distance? You have it starting out at about 170 fps based on your chart. The drop of 6.8 feet occurs with the fragment starting out on a horizontal or low upward angle. If it was a horizontal initial trajectory, that means it took t=√(2x6.8/32)=.65 seconds to travel that 141 feet so its average speed was 141/.65=216 fps. Pretty hard to average 216 fps when you start out at 170 fps and then keep slowing down.
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Only one of you guys can be right, but both could be wrong. The abundance of assumptions is staggering. Fascinating to watch two LNs battle it out!
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Only one of you guys can be right, but both could be wrong. The abundance of assumptions is staggering. Fascinating to watch two LNs battle it out!
It's almost as much fun as watching two zombified-by-KGB*-disinformation JFKA CT's squabble over their tinfoil-hat theories!
*Today's SVR and FSB
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Yes. But the time it takes to add a distance h is ½gt2 so for an additional 4 feet, t=√(2x4/32)=.5 seconds. That is how long it takes for the fragment to travel from the place where it strikes the asphalt to where it strikes the manhole (in your scenario). So its average speed over that 250 feet is 250/.5=500 fps.
If the fragment:
- started out at 1500 fps, meaning the bullet would have lost only 44% of its incident energy of 1860 J (assuming the bullet struck asphalt at 2000 fps) - which still works out to 800 Joules - and
- averaged 500 fps over that 250 ft distance, and
- was able to drop a further 6.8 feet over the next 141 feet
what do you say was its speed over the last 141 ft distance? You have it starting out at about 170 fps based on your chart. The drop of 6.8 feet occurs with the fragment starting out on a horizontal or low upward angle. If it was a horizontal initial trajectory, that means it took t=√(2x6.8/32)=.65 seconds to travel that 141 feet so its average speed was 141/.65=216 fps. Pretty hard to average 216 fps when you start out at 170 fps and then keep slowing down.
Now, you seem to be discussing the second ricochet (from the manhole cover apron to the curb on the south side of Main Street). We will never know the specific velocities and angles for certain. So, as I said before we have to assume some reasonable numbers. The 1500-fps velocity figure for the first ricochet is what seemed to me to be a reasonable guesstimate before I started my own experimentation. My experiments with 30-degrees ricochets have them retaining a much higher percentage of their velocities than that, closer to 95%. So this might surprise you but when we increase the velocity of the first ricochet to 1900-fps we get these results. These are within a reasonable range for this discussion, especially since we are not using exact measurements and only reasonable assumptions, etc.
(https://i.vgy.me/FkK1el.png)
(https://i.vgy.me/PLWYz8.png)
For the second ricochet, if we assume an angle of positive 1-degree, simple geometry tells us the drop needs to increase by ~29.5” from ~81.6” to a total of ~111.1”. And if we a assume a velocity of ~187.8-fps, the fragment fragmented again and is now more aerodynamic, we get these results:
(https://i.vgy.me/pnggAH.png)
(https://i.vgy.me/iX1wKi.png)
Again, well within a range suitable for this discussion.