Determine vertical velocity vector on sloped surface

AI Thread Summary
The discussion revolves around determining the vertical and horizontal velocity components of a golf ball launched on a sloped surface. The vertical velocity is calculated using the formula v_{f,v} = v_f * cos(β) * sin(φ), while the horizontal velocity is derived differently, leading to confusion. Participants question why the author does not apply the same logic for the horizontal component by multiplying with cos(φ). The conversation highlights the importance of considering the three-dimensional nature of the problem and how the slope affects the velocity components. Clarification is sought on the correct formulation, especially when both x-axis and y-axis slopes are involved.
hjam24
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A golf is launched at a speed v,f and launch angle, β,f. The slope of the green is equal to φ. At some point the ball is located on the rim of a hole. The side view (a) and overhead view (b) looks as in the attached image.According to the author of the [paper][2] "The Physics of Putting" the vertical velocity is given by

$$ v_{f,v} = v_f \cdot \cos(\beta) \cdot \sin(\phi) $$

The horizontal velocity is equal to:

$$ v_{f,h} = v_f \cdot (\cos^2(\beta_f) \cos^2(\phi) + \sin^2(\beta_f))^{1/2} $$

I understand the reasoning behind the vertical velocity component. Basically we have the hypotenuse, v,fy (v,f * cos(β,f)) and multiply this with sin(φ). Now we have the opposite. However, why is the author not following the same logic for the horizontal velocity and multiplying the hypotenuse with cos(φ)? Please show me how his formulation is derived.

My (unsuccessful) attempt:

Pythagorean theorem

$$ \sin^2(\phi) + \cos^2(\phi) = \frac {v_{f,h}^2 + v_{f_v}^2} {v_f^2} $$

$$ v_f^2 \cdot (\sin^2(\phi) + \cos^2(\phi)) = v_{f,h}^2 + v_{f_v}^2 $$

$$ v_{f,h}^2 = v_f^2 \cdot (\sin^2(\phi) + \cos^2(\phi)) - v_{f_v}^2 $$

Substitute vertical velocity according to author

$$ v_{f,h}^2 = v_f^2 \cdot (\sin^2(\phi) + \cos^2(\phi)) - (v_f\cos(\beta)\sin(\phi))^2 $$

v,f is v,fy

$$ v_{f,h}^2 = (v_f\cos(\beta))^2 \cdot (\sin^2(\phi) + \cos^2(\phi)) - (v_f\cos(\beta)\sin(\phi))^2 $$

$$ v_{f,h}^2 = v_f^2\cos^2(\beta) \cdot (\sin^2(\phi) + \cos^2(\phi)) - v_f^2\cos^2(\beta)\sin^2(\phi) $$

$$ v_{f,h}^2 = v_f^2\cos^2(\beta)\sin^2(\phi) + v_f^2\cos^2(\beta)\cos^2(\phi) - v_f^2\cos^2(\beta)\sin^2(\phi) $$

$$ v_{f,h}^2 = v_f^2\cos^2(\beta)\cos^2(\phi) $$

$$ v_{f,h}^2 = v_f^2\cos^2(\beta)\cos^2(\phi) $$

$$ v_{f,h} = v_f(\cos^2(\beta)\cos^2(\phi))^{1/2} $$

Let me know what I am doing wrong. Thanks in advance

[2]: http://www.puttingzone.com/Science/cjp-putting.pdf
 

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After getting the formula of ##v_{f\ vertical}##, the author seems to have made use of Pythagoras relation
v_{f\ vertial}^2+v_{f\ horizontal}^2=v_f^2
So
v_{f\ horizontal}^2=v_f^2-v_{f\ vertial}^2
 
You have to consider all three dimensions. Your logic applies to the y- and z- components, but it neglects the x-component. To see this, consider the case where ##\phi=0##. Using your logic, you'd say ##v_{f,v}=0##, as expected, but ##v_{f,h}=v_f\cos\beta##. But if there's no vertical component, shouldn't ##v_{f,h}=v_f##?

A straightforward way to see what's happening is to first consider the case where ##\phi=0##. Then the components of the velocity will be
$$\mathbf v_f = -v_f \sin\beta\,\mathbf i + v_f \cos\beta\,\mathbf j + 0 \,\mathbf k.$$ Now add in the effect of the uphill slope, which mathematically is a rotation by an angle ##\phi## about the x-axis. The rotation will leave the x-component unchanged but will turn the y-component into a combination of y- and z- components.
 
Poster has been reminded not to generate multiple thread starts on the same question (2 threads merged)
A golf ball is launched at a speed v and launch angle, β . At some point the ball is located on the rim of a hole. The view from above looks as follows:

top.png


Vector v can be expressed in two components, v_v and v_y:$$v_y = v \cdot \cos(\beta)$$

and

$$v_x = v \cdot \sin(\beta)$$

If we look at the surface from a different angle, we see that the surface has a slope.
The slope of the green is equal to φ (along the y-axis).

side.png

v_v is perpendicular to the y-axis. As a result:

$$v_v = v \cdot \cos(\beta) \cdot \sin(\varphi)$$

Let's assume now there is no slope across the y-axis (φ=0), but there is a slope across the x-axis, θ.

front.png


Following the same logic v_v is perpendicular to the x-axis. As a result:

$$v_v = -1 \cdot v \cdot \sin(\beta) \cdot \sin(\theta)$$

What is the formula of v_v expressed in terms of both φ and θ? So, in the case that the surface is sloped/inclined across both the x- and y-axis?

[Moderator's note: moved from a technical forum.]
 
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hjam24 said:
A golf ball is launched at a speed v and launch angle, β . At some point the ball is located on the rim of a hole. The view from above looks as follows:
@hjam24 -- Why did you repost this question and show no work? Were the replies above to your original thread a couple weeks ago not helpful enough?
 
berkeman said:
@hjam24 -- Why did you repost this question and show no work? Were the replies above to your original thread a couple weeks ago not helpful enough?
Hi,

The previous question was with a singe slope across one axis, this question involves a slope across bot the x-axis and the y-axis. I believe the difference is large enough to open a new topic. I am sorry if I misjudged that
 
Well the wording was so similar that it was hard to tell the difference.

hjam24 said:
What is the formula of v_v expressed in terms of both φ and θ? So, in the case that the surface is sloped/inclined across both the x- and y-axis?
Please show your work on this, and you should continue to get good help. Thank you.
 
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hjam24 said:
why is the author not following the same logic for the horizontal velocity and multiplying the hypotenuse with cos(φ)?
In the direction of the ball, the slope is not ##\phi##.
hjam24 said:
$$ v_{f,h}^2 = v_f^2 \cdot (\sin^2(\phi) + \cos^2(\phi)) - (v_f\cos(\beta)\sin(\phi))^2 $$

v,f is v,fy

$$ v_{f,h}^2 = (v_f\cos(\beta))^2 \cdot (\sin^2(\phi) + \cos^2(\phi)) - (v_f\cos(\beta)\sin(\phi))^2$$
I don’t understand what you did there. You seem to have just multiplied a term by ##\cos^2(\beta)##.
 
hjam24 said:
A golf ball is launched at a speed v and launch angle, β . At some point the ball is located on the rim of a hole.
What is your question for the second case?
The vertical velocity vector is perpendicular to both top views, as well as to the axis that is not tilted (or remains horizontal).
 
  • #10
It would be nice if the OP actually replied to the posts so we know if he understood what was said or not.
 
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