• May 11th 2007, 09:49 PM
daewoo_lowrider

Suppose a try is scored at T, then the kicker can kick to the posts from anywhere along the line TF perpendicular to the try line PQ. (Refer to Figure 1) Where is the best place to kick from?
Many kickers seem to think that the further they go back (within their kicking limit) the better. But this is not necessarily the case. The best position for the kick to be taken is clearly where angle ACB subtended by the posts along TF is a maximum.

Here is the diagram:
http://img.photobucket.com/albums/v3...ce/figure1.jpg
Basically, I think I have to use trig, and a circle subtending the goal posts, but I'm stuck at that, I have hardly any measurements and angles (the distance between the goal posts is 5.6m.

Thanks for any help,
Daniel
• May 12th 2007, 04:14 AM
Geometor
what's the question? :S
• May 12th 2007, 04:20 AM
daewoo_lowrider
Where is the best place to kick from?

I think I figured out how to do it, basically assume the try is scored 1m away from goals and use trig to find the angle 1m back, 2m back etc. Then I assume the try was scored 2m away and so on.

Daniel
• May 12th 2007, 05:20 AM
earboth
Quote:

Originally Posted by daewoo_lowrider
Where is the best place to kick from?

I think I figured out how to do it, basically assume the try is scored 1m away from goals and use trig to find the angle 1m back, 2m back etc. Then I assume the try was scored 2m away and so on.

Daniel

Hello, Daniel,

In this case you must have used a constant distance AT. Which value has this distance and why?

If I know this value I can offer you some equations for a start.

Are you supposed to use calculus to find the maximum value of the angle?
• May 12th 2007, 05:26 AM
daewoo_lowrider
Hi earboth, sorry if I was unclear. I was not given any dimensions, thats why i assumed you make up a series of measurements for AT (I have done 1-5m). And then I calculated the angle of C by using Tan, counting back each meter until the angle reaches a maximum. I then recorded these coordinates.

I was then going to use these coordinates to calculate an equation for a hyperbola - the arc of this hyperbola would then show the best position for the player to kick across the segment of the line PA.

Does this sound correct?
Thanks,
Daniel
• May 12th 2007, 05:40 AM
Soroban
Hello, Daniel!

Quote:

The kicker can kick from anywhere along the line TF (see diagram).
Where is the best place to kick from?

Code:

T  d  A      5.6      B
* - - - * - - - - - - - *
|      *            *
|    *          *
x |    *        *
|  *      *
|β *    *
| * θ *
|* *
C *
|
F *

Let TA = d, .TC = x.
Let /ACB = θ, ./TCA = β

We want to maximize θ.
. . Note that: .tanβ = d/x .[1]

. . . . . . . . . . . . . . . . . . d + 5.6
We have: .tan(θ + β) .= .---------
. . . . . . . . . . . . . . . . . . . . x

. . . . . . .tanθ + tanβ . . . . d + 5.6
Then: . ----------------- .= . ----------
. . . . . .1 + tanθ·tanβ . . . . . .x

. . . . . . . . . . . . .tanθ + (d/x) . . . d + 5.6
Substitute [1]: . ----------------- .= .---------
. . . . . . . . . . . .1 + (d/x)·tanθ . . . . .x

. . . . . . . . . . . . . . . . . . . . . . .-5.6x
This simplifies to: .tanθ .= .------------------
. . . . . . . . . . . . . . . . . . . .x² + d² + 5.6d

. . . . . . . . . . . . . . . . . . . . . . . . . . . .d² + 5.6d - x²
Differentiate: .sec²θ·(dθ/dx) .= .-5.6 [---------------------]
. . . . . . . . . . . . . . . . . . . . . . . . . . .(x² + d² + 5.6d)²

The derivative equals zero when that numerator equals zero.

Hence: . d² + 5.6d - x² .= .0 . . . . .= .d² + 5.6d
. . . . . . . . . . . . . ________
Therefore: .x .= .√d² + 5.6d

• May 12th 2007, 06:25 AM
earboth
Quote:

Originally Posted by daewoo_lowrider
...

I was then going to use these coordinates to calculate an equation for a hyperbola - the arc of this hyperbola would then show the best position for the player to kick across the segment of the line PA
....

Hi, Daniel,

I first tried your method to solve the problem too but I didn't get a plausible result. So I started again and was looking for a geometric solution.

I believe that I found one but :rolleyes: .. I'm not sure but maybe you can use my bits and pieces to get an accurate solution.

EDIT: Don't believe me. Have a look at Soroban's solution. You can use my result as a rough approximation only.
• Dec 7th 2007, 06:28 PM
mr fantastic
Calculus unnecessary - easy geometric solution
Quote:

Originally Posted by Soroban
Hello, Daniel!

Code:

T  d  A      5.6      B
* - - - * - - - - - - - *
|      *            *
|    *          *
x |    *        *
|  *      *
|β *    *
| * θ *
|* *
C *
|
F *

Let TA = d, .TC = x.
Let /ACB = θ, ./TCA = β

We want to maximize θ.
. . Note that: .tanβ = d/x .[1]

. . . . . . . . . . . . . . . . . . d + 5.6
We have: .tan(θ + β) .= .---------
. . . . . . . . . . . . . . . . . . . . x

. . . . . . .tanθ + tanβ . . . . d + 5.6
Then: . ----------------- .= . ----------
. . . . . .1 + tanθ·tanβ . . . . . .x

. . . . . . . . . . . . .tanθ + (d/x) . . . d + 5.6
Substitute [1]: . ----------------- .= .---------
. . . . . . . . . . . .1 + (d/x)·tanθ . . . . .x

. . . . . . . . . . . . . . . . . . . . . . .-5.6x
This simplifies to: .tanθ .= .------------------
. . . . . . . . . . . . . . . . . . . .x² + d² + 5.6d

. . . . . . . . . . . . . . . . . . . . . . . . . . . .d² + 5.6d - x²
Differentiate: .sec²θ·(dθ/dx) .= .-5.6 [---------------------]
. . . . . . . . . . . . . . . . . . . . . . . . . . .(x² + d² + 5.6d)²

The derivative equals zero when that numerator equals zero.

Hence: . d² + 5.6d - x² .= .0 . . . . .= .d² + 5.6d
. . . . . . . . . . . . . ________
Therefore: .x .= .√d² + 5.6d

Or ....... you could recognise that the spot you want will be the point where the circle passing through A and B is >tangent< to the line TF. This follows from a couple of simple circle geometry theorems. Then some trivial algebra to get the distance from the tryline.

By the way ... just passing by and couldn't resist posting the >simple< approach.

Ciao.
• Feb 26th 2017, 10:08 PM
nurininsyirah97