Warrior General Games

Originally posted by SyedAshrafulla
It's definitely way dependent on the velocity magnitude v, because the goal posts are 10 feet high (so the time of flight is more confusing). Normally in your 45-degree estimate, it's b/c v t sin Θ - 1/2 a t^2 = 0, so t = 2v/a sin Θ and thus the distance covered is vt cos Θ = v^2/a sin 2Θ which maximizes over Θ when 2Θ = π/2, or Θ = π/4 (45 degrees).

The problem is that now you constrain v t sin Θ - 1/2 a t^2 = 10, which means the ball meets the plane of the uprights at a much more complicated expression: t = (v sin Θ + sqrt(v^2 sin^2 Θ - 20g))/a and thus we want to maximize, over Θ, the function d(Θ = v/a cos Θ (v sin Θ + sqrt(v^2 sin^2 Θ - 20g)) = v/a (v/2 sin 2Θ + cos Θ sqrt(v^2 sin^2 Θ - 20g)). Notice that if we were dealing with no crossbar, the 20g would go away and we'd get the above solution. We can ignore the v/a too.

So the summary is that the optimal Θ is one that maximizes v/2 sin 2Θ + cos Θ sqrt(v^2 sin^2 Θ - 20g)). I tried to tkae the derivative, etc. and it turns out to be a cubic equation in the variable sin^2 Θ, which means it's solvable and probably only has one solution in [0, π/2].

wat

Originally posted by SyedAshrafulla
It's definitely way dependent on the velocity magnitude v, because the goal posts are 10 feet high (so the time of flight is more confusing). Normally in your 45-degree estimate, it's b/c v t sin Θ - 1/2 a t^2 = 0, so t = 2v/a sin Θ and thus the distance covered is vt cos Θ = v^2/a sin 2Θ which maximizes over Θ when 2Θ = π/2, or Θ = π/4 (45 degrees).

The problem is that now you constrain v t sin Θ - 1/2 a t^2 = 10, which means the ball meets the plane of the uprights at a much more complicated expression: t = (v sin Θ + sqrt(v^2 sin^2 Θ - 20g))/a and thus we want to maximize, over Θ, the function d(Θ = v/a cos Θ (v sin Θ + sqrt(v^2 sin^2 Θ - 20g)) = v/a (v/2 sin 2Θ + cos Θ sqrt(v^2 sin^2 Θ - 20g)). Notice that if we were dealing with no crossbar, the 20g would go away and we'd get the above solution. We can ignore the v/a too.

So the summary is that the optimal Θ is one that maximizes v/2 sin 2Θ + cos Θ sqrt(v^2 sin^2 Θ - 20g)). I tried to tkae the derivative, etc. and it turns out to be a cubic equation in the variable sin^2 Θ, which means it's solvable and probably only has one solution in [0, π/2].
Solving the above equation will give you the angle to kick in order to hit the cross bar


Yes, the basic kinematic formula hold true. The question isn't in the kinematics of the problem, y>10ft. the question is the transfer of momentum to the ball. If you assume the initial velocity is independent of the initial angle, then 45 will be the best you can do. But imagine kicking the ball at an 80 degree angle. Don't think you could get same velocity as say 30 degrees.

So in order to solve for the optimal angle, you would first have to find the equation to solve for the initial velocity based on where and how the kicker kicked the ball. My guess is that it would a form of impulse momentum equation, but then you got to take in account the give of the ball in order to calculate the impulse side of the equation. It would be slightly more difficult than physics I. To get close enough to reality, it would be a grad student project involving 3D nonlinear dynamics, and some mechanic of materials.



Edit: Conclusion
Option 1: Kicker only kick one angle-- best angle =45
Option 2: Kicker can kick multiple angles: velocity must vary with angle(not sure how yet) best angle= ? based on V=f(theta)

Originally posted by Firenze

wat

My thoughts exactly.

Originally posted by shuebru
Originally posted by SyedAshrafulla

It's definitely way dependent on the velocity magnitude v, because the goal posts are 10 feet high (so the time of flight is more confusing). Normally in your 45-degree estimate, it's b/c v t sin Θ - 1/2 a t^2 = 0, so t = 2v/a sin Θ and thus the distance covered is vt cos Θ = v^2/a sin 2Θ which maximizes over Θ when 2Θ = π/2, or Θ = π/4 (45 degrees).

The problem is that now you constrain v t sin Θ - 1/2 a t^2 = 10, which means the ball meets the plane of the uprights at a much more complicated expression: t = (v sin Θ + sqrt(v^2 sin^2 Θ - 20g))/a and thus we want to maximize, over Θ, the function d(Θ = v/a cos Θ (v sin Θ + sqrt(v^2 sin^2 Θ - 20g)) = v/a (v/2 sin 2Θ + cos Θ sqrt(v^2 sin^2 Θ - 20g)). Notice that if we were dealing with no crossbar, the 20g would go away and we'd get the above solution. We can ignore the v/a too.

So the summary is that the optimal Θ is one that maximizes v/2 sin 2Θ + cos Θ sqrt(v^2 sin^2 Θ - 20g)). I tried to tkae the derivative, etc. and it turns out to be a cubic equation in the variable sin^2 Θ, which means it's solvable and probably only has one solution in [0, π/2].
Solving the above equation will give you the angle to kick in order to hit the cross bar


Yes, the basic kinematic formula hold true. The question isn't in the kinematics of the problem, y>10ft. the question is the transfer of momentum to the ball. If you assume the initial velocity is independent of the initial angle, then 45 will be the best you can do. But imagine kicking the ball at an 80 degree angle. Don't think you could get same velocity as say 30 degrees.

So in order to solve for the optimal angle, you would first have to find the equation to solve for the initial velocity based on where and how the kicker kicked the ball. My guess is that it would a form of impulse momentum equation, but then you got to take in account the give of the ball in order to calculate the impulse side of the equation. It would be slightly more difficult than physics I. To get close enough to reality, it would be a grad student project involving 3D nonlinear dynamics, and some mechanic of materials.



Edit: Conclusion
Option 1: Kicker only kick one angle-- best angle =45
Option 2: Kicker can kick multiple angles: velocity must vary with angle(not sure how yet) best angle= ? based on V=f(theta)45 is not a value that maximizes that function, because its derivative at 45 is 1 + sqrt((v^2 - 40g)/4), which is definitely not 0. That immediately shows even if hte kicker can only kick one angle, 45 is not the answer for a fixed velocity v.

Ok, your math is correct after I looked into it. I thought you were some wanna be hot shot dude that just learned the kinematic formulas. After I seen that the first time, I quit reading.
I used matlab to solve it out and give a plot vs velocity. It started off complex(which makes sense because low velocities the ball will never reach 10 ft) and around 28ft/s the angle became real and very close to 90 degrees. Then as velocity approaches infinity, the angle approaches 45.

The question now is why do professionals kick it at roughly 35 like the OP said, and that goes back to my answer of it isn't the kinematics of the ball, but the kinetics.

Originally posted by shuebru
Ok, your math is correct after I looked into it. I thought you were some wanna be hot shot dude that just learned the kinematic formulas. After I seen that the first time, I quit reading.
I used matlab to solve it out and give a plot vs velocity. It started off complex(which makes sense because low velocities the ball will never reach 10 ft) and around 28ft/s the angle became real and very close to 90 degrees. Then as velocity approaches infinity, the angle approaches 45.

The question now is why do professionals kick it at roughly 35 like the OP said, and that goes back to my answer of it isn't the kinematics of the ball, but the kinetics.
I totally agree with your last statement, but I have a feeling that without the simplicity, it will be hard to run all the FG simulations quickly on the server. I mean, 5 FG att/game with anything more than kinematics might require a bazillion lookup tables.

It makes sense that it's asymptotic to 45 degrees (at infinity 10 feet high looks like 0 feet high). Interesting that you have to kick the ball at least 28 ft/s to get lucky, I guess I just never thought it that far out (my copy of MATLAB was running an overnight sim). Pretty neat though! Can you post a hires version of the graph as a picture (code would be even better!) so we can see how the curve looks?

It's people like you two that will make this game a HELL of a lot more realistic

The physics logic in the 1st post is a little off but overall good idea

Originally posted by Firenze
Originally posted by SyedAshrafulla

It's definitely way dependent on the velocity magnitude v, because the goal posts are 10 feet high (so the time of flight is more confusing). Normally in your 45-degree estimate, it's b/c v t sin È - 1/2 a t^2 = 0, so t = 2v/a sin È and thus the distance covered is vt cos È = v^2/a sin 2È which maximizes over È when 2È = ð/2, or È = ð/4 (45 degrees).

The problem is that now you constrain v t sin È - 1/2 a t^2 = 10, which means the ball meets the plane of the uprights at a much more complicated expression: t = (v sin È + sqrt(v^2 sin^2 È - 20g))/a and thus we want to maximize, over È, the function d(È = v/a cos È (v sin È + sqrt(v^2 sin^2 È - 20g)) = v/a (v/2 sin 2È + cos È sqrt(v^2 sin^2 È - 20g)). Notice that if we were dealing with no crossbar, the 20g would go away and we'd get the above solution. We can ignore the v/a too.

So the summary is that the optimal È is one that maximizes v/2 sin 2È + cos È sqrt(v^2 sin^2 È - 20g)). I tried to tkae the derivative, etc. and it turns out to be a cubic equation in the variable sin^2 È, which means it's solvable and probably only has one solution in [0, ð/2].

wat

This is EPIC

Originally posted by Nuge20
The physics logic in the 1st post is a little off but overall good idea

The physics logic is straight from the mouth of an NFL kicker.

Originally posted by SyedAshrafulla
It's definitely way dependent on the velocity magnitude v, because the goal posts are 10 feet high (so the time of flight is more confusing). Normally in your 45-degree estimate, it's b/c v t sin Θ - 1/2 a t^2 = 0, so t = 2v/a sin Θ and thus the distance covered is vt cos Θ = v^2/a sin 2Θ which maximizes over Θ when 2Θ = π/2, or Θ = π/4 (45 degrees).

The problem is that now you constrain v t sin Θ - 1/2 a t^2 = 10, which means the ball meets the plane of the uprights at a much more complicated expression: t = (v sin Θ + sqrt(v^2 sin^2 Θ - 20g))/a and thus we want to maximize, over Θ, the function d(Θ = v/a cos Θ (v sin Θ + sqrt(v^2 sin^2 Θ - 20g)) = v/a (v/2 sin 2Θ + cos Θ sqrt(v^2 sin^2 Θ - 20g)). Notice that if we were dealing with no crossbar, the 20g would go away and we'd get the above solution. We can ignore the v/a too.

So the summary is that the optimal Θ is one that maximizes v/2 sin 2Θ + cos Θ sqrt(v^2 sin^2 Θ - 20g)). I tried to tkae the derivative, etc. and it turns out to be a cubic equation in the variable sin^2 Θ, which means it's solvable and probably only has one solution in [0, π/2].

i see... but if bort messes up the sqrt(v^2 sin^2 Θ - 20g), then the Angle Projectory will only be v/2 sin 2Θ - cos Θ ending in the end of the world as we know it.

wow, it cant bold the minus sign very well =/

Lotta too-smart-for-me talk goin on in here...I'm running away.

oh by the way, this analysis (all of it) goes out the window if there is no crossbar simulated in GLB field goals. if it's simulated that you just have to cross the plane within the sticks (Australian football rules) then 45 is 45. makes for a ton of changes to how we think about kicking though (like making strength as useless as it apparently already is).

Originally posted by SyedAshrafulla
oh by the way, this analysis (all of it) goes out the window if there is no crossbar simulated in GLB field goals. if it's simulated that you just have to cross the plane within the sticks (Australian football rules) then 45 is 45. makes for a ton of changes to how we think about kicking though (like making strength as useless as it apparently already is).

The math involved assumes you gotta get over the crossbar, which means you have to be able to kick farther than exactly to the posts. It's not super complicated, like I said, though. Air resistance, wind, friction, transfer of momentum, etc, are all ignored. Get too complicated and it takes too long to make calculations.