skouperd's balls

[flycatchr]

Ps, Flycatchr, if you are really interested in the topic, then perhaps you could share your views with me with another mathematical problem I've been having for several years relating to the optimal angle to launch an objet thus achieving maximum distance. In plain english, given a person' club swing speed at moment of impact, what is the optimal angle at which to launch a golf ball for maximum distance. I wrote a piece on this many years ago but never got a sufficient, cool, nice looking equation out of it. The guys I was corresponding then failed to grasp the concept and remained adament that the answer is 45 degrees (which equates to around an 8 iron) and obviously wrong.

the figure "45 degrees" is the optimal launch angle for a projectile to get the greatest distance. IIRC this result is from a mathematical formula who's name i forget atm. (maximise for angle)
what the formula, and your nana friend fail to take into account is a commen problem in the academic world - REAL LIFE.

more to follow, as REAL life has hit me

[Megageth]

A parabola of 45 degrees provides the most distance in theory but this will not apply to golf. As you say, an 8 iron is probably the closest club to 45 degrees but the impact is not clean. Much of the impact on tilted irons is downward over the surface of the ball rather than through the centre of the ball's mass in a direct line of its initial vector (before gravity and resistance alter its path). The 45 degree angle also fails to take into account the bounce of the ball. As a result, 'woods' that are angled closer to the ball's initial vector (dependant on swing dynamics) will gain further distance.

[flycatchr]

the biggest problem with the 45 degree hypothesis and golf is that the direction of swing, and therefore force, IS NOT 45 degrees, it is ZERO degrees (also depending on swing characteristics)
i am sure that if we could hit that ball AT 45 degrees, with some sort of accuracy, we may have a solution.

here is my thinking so far:
lets start at the beginning and define the variables

assumptions and exclusions:
A. we have the height of the golfer, and his own club head speed - assume this to be constant.
B. club length relative to the golfer - assume to be fixed and not a problem
C. assume contact with the ball to be consistent for every strike (pro golfer)
D. we will ignore the effects of shaft inconsistencies/bad manufacture
E. the ball is stroked in the sweet spot every time (tiger's dream )
F. side winds can be ignored
G. as well as side spin for this argument
H. THUS we have a 2D equation, NOT a 3d one
I. the placement of the ball doesn't change the angle off attack of the club head - IE the stated club head angle is the angle it strikes the ball.
J. We could for the sake of this argument also have the club head speed consistent for each set of equations.
K. AND keep the density of the air constant - perhaps sea level.

Possible inconsistencies in this argument:
I. Different clubs manufactures will have varied grip on the ball at moment of impact
II. Different balls will have varied grip on the club at moment of impact
III. Different balls will have different dimples, therefore the flight path will differ
IV. Different balls have different bounce characteristics
V. The first bounce is critical and the resultant roll distance depends on ground conditions. (rain, mud, ground hardness, ball slip off grass etc)
VI. Scratches and nicks on the ball.
VII. Each golfer has a different flight impact characteristic, depending on his swing, grip, fatigue (how many mistresses he has), state of mind and shoe comfort.


so we are left with:
1. Angle of club head
2. Club head speed
3. Amount of backspin on ball

[Skouperd]

Here was the article I wrote on my theories many, many moons ago....

http://www.skoups.com/_articles/golf_balls.html

The synopsis (ignore the grammar and spelling the forum I posted this on originally did not care about either):

Gravity always exists, and it will start pulling the ball back to earth as soon as the ball leave the tee. But the velocity that the ball is moving forward and the upward motion is greater than the velocity of the earth's gravity. It therefor means that the force gravity is playing on the ball is virtually zero. The 2 biggest forces are off course the forward motion on the ball as well as the air friction reducing the forward motion. (Something like if you have two direct forces against each other, they will have the greatest impact. More or less a head to head accident cause a bigger damage than a side impact accident) Therefor gravity will only really start to come into effect when the forward motion less air friction is LESS than the gravitational pull on the ball. (I have no clue how to say this simpler)

But as soon as gravity starts to kick in, the ball is already say 100 meters in the air. It will still increase height for a few meters but then it will start to reduce its forward (and upward) motion and eventually come back to earth. Yes at this stage the angle of the ball coming down would be the same, (i.e. 9.5') BUT only until it reaches **the mark where the difference in forward motion less air friction is LESS than the force of gravity. (I.e. the 100 meter in the air)

But seeing that the air friction still plays a large part on the forward motion of the ball, the ball still reduces speed. When we get to the 100-meter mark in the air (** refer back to the above paragraph) the forward motion reduce much further and the effect gravity have on the ball is greater. That is why the ball will not be coming down at the same angle as it started but as it increases it's deceleration the forward motion because virtually null. (I.e. when the ball fall below the 100-meter height mark)

This is the reason why the ball seems to follow the path, which I described in a previous message.

Does this mean that if we have a lower trajectory we will hit a golf ball longer? No. Why not? The upward force the ball is exercising might not be enough when the forward motion less the air friction is LESS than the gravitational pull. This will cause that the ball start falling at a very low height i.e. 30 or 50 meters in stead of the 100 meters used in the previous example. So then even though the forward motion originally was the same as the first example, the height the "swap" occur will give you that extra few yards.

So to come back to my original question, what loft should I use? It would be where one gets the maximum height at the place where the forward motion combined with the height motion is equal to the gravity plus the air friction.

So based on the above I am not so sure that we can in fact use the law of Newton here. I am not sure if Newton's law actually cater for horizontal speeds that exceed the 9.8/s.s barrier. Based on the above (proven golf ball flight.) we actually have several factors that do come into play. They are as follows:

Gravity (bring the ball back to earth) Forward motion (the energy applied to the ball when you hit it) Air Friction (the friction that the ball causes by moving forward) Upwards motion (the trajectory of the ball at point of impact)

(Then off course you get your fade and draw motions as well, but I think the above is complex enough to try and explain)

Based on the above I believe it is possible to say to a person with what degree driver (the number 1 knop-kierrie) he must play with in order to get the maximum distance by ONLY taking into consideration the amount of energy he is able to release at impact.

So now I know why I love golf….

Adios
Skouperd

[flycatchr]

technically speaking skoups, gravity is always acting on the ball, but ONLY down, wheras drag is directly opposite to the direction of travel.
the real trick here is to know that the flatter the trajectory, the faster the initial velocity is, BUT, there is a maximum angle of attack (resulting in trajectory) that will privide the greatest distance, and that is what you are asking for

[gr@v1ty]

Gravity always exists

That's all that matters...

[flycatchr]

Gravity always exists

That's all that matters...

relatively speaking, maybe

[XennoX]

Thread necro.

Gravity, being considered a vertical force, will have NO influence over the horizontal velocity and distance. You cannot add orthogonal vectors to each other as their influence on the other is zero.

The equation(s) you would have to develop would be the trajectory/ballistic equation, but you would have to add in the angular momentum of the golf ball and the drag force on the ball.

FD = 0.5­·CDrag·Dfluid·AProjected·v², where:

FD is the Drag Force;
CDrag is the Coefficient of Drag;
Dfluid is the density of the fluid;
AProjected is the projected area of the object;
v is the velocity of the object.

Please note, Force is a vector and so if Velocity, the rest are all scalars.

Angular Momentum (aka Backspin) is given by the equation:

L = I·w

Angular momentum is a Vector, known as a moment. (Like Torque and Bending)

L represents the quantity for the angular momentum
I is the moment of inertia of the object
w is the angular velocity, often given as Radians Per Second (Rad/s)

The units of Angular Momentum are N·m·s as opposed to N·m for Torque and Bending

Your equation would also involve the use of Calculus, to find the maxima you would have to differentiate your function, but you would also probably have to integrate to find the total displacement.

Newton's equation of motion F = ma CAN be used, as this is a Newtonian problem. (Note I bold the quantities that are vectors)

It also may be easier to use a polar co-ordinate system as opposed to a cartesian co-ordinate system.

So it would look something like this:

Vertical Displacement = Initial Vertical Velocity - Gravitational Acceleration + The Lift created due to the Magnus Effect (aka Backspin) - Vertical Drag
Horizontal Displacement = Initial Horizontal Velocity - Horizontal Drag

[Skouperd]

Xennox, I observe you've been studying, did you pass?

[XennoX]

Almost, one more year. One more year and them I'm done.