Determine height difference between start and end point on inclined plane

[HJAM24]

Hi all,

I have a question regarding the below figure. It represents an inclined slope (theta) along the x-axis.
A ball moves from point A to C.

As a result:

[math]AC_x = AC \cdot \sin(\beta)[/math]
and

[math]AC_y = AC \cdot \cos(\beta)[/math]

Line AB is the perpendicular line of the two parallel line with slope theta. This means that the height of point A is the same as the height of point B.
How do I determine the vertical distance between point A and point C?




A

 

[HJAM24]

A new point should be drawn, let's say point D. BCD will form a triangle. The vertical distance would be line BD.
I assume / believe that BD can be expressed in terms of AB.
Basically I want to recreate Figure 8a and formula 27b of this paper

 

[Deleted member 4993]

Hi all,

I have a question regarding the below figure. It represents an inclined slope (theta) along the x-axis.
A ball moves from point A to C.

As a result:

[math]AC_x = AC \cdot \sin(\beta)[/math]
and

[math]AC_y = AC \cdot \cos(\beta)[/math]

Line AB is the perpendicular line of the two parallel line with slope theta. This means that the height of point A is the same as the height of point B.
How do I determine the vertical distance between point A and point C?


View attachment 33507

A

By "vertical" distance do you mean:

Distance parallel to y-axis
Or
Distance perpendicular to the ground

 

[HJAM24]

I mean distance perpendicular to the ground

 

[Cubist]

[math]AC_x = AC \cdot \sin(\beta)[/math]

You're missing a negative, as drawn, the x component of vector AC is [imath]AC_x = - AC \cdot \sin(\beta)[/imath]

Line AB is the perpendicular line of the two parallel line with slope theta. This means that the height of point A is the same as the height of point B.

Does this mean that your figure has depth or in other words each point has (potentially) different z coordinate?

 

[HJAM24]

You're missing a negative, as drawn, the x component of vector AC is ACx=−AC⋅sin⁡(β)AC_x = - AC \cdot \sin(\beta)ACx=−AC⋅sin(β)

Okay, and if AC would be on the right of the y-axis the formula is without a minus sign?

Does this mean that your figure has depth or in other words each point has (potentially) different z coordinate?

Yes, I showed the situation as as XY plane but it is actually a xyz plane. My figure is a modification of the below figure. The below figure shows no slope across the x-axis but a slope across the y-axis.

 

[Cubist]

Okay, and if AC would be on the right of the y-axis the formula is without a minus sign?

Correct

Yes, I showed the situation as as XY plane but it is actually a xyz plane. My figure is a modification of the below figure. The below figure shows no slope across the x-axis but a slope across the y-axis.

View attachment 33510

I can now see your second post (we had to wait for a moderator to approve it).

I'm immediately put off from reading any of that paper because figure 8 doesn't actually show "a) side view" and "b) overhead view" of the same thing! In the side view the golf ball is teetering on the edge of the hole. In the overhead view the golf ball is distance R back from the edge.

Have you read the paper? Do the diagrams make sense to you? (And if so then can you explain please since I feel a bit )

 

[HJAM24]

Correct

Makes sense.

In the side view the golf ball is teetering on the edge of the hole. In the overhead view the golf ball is distance R back from the edge.

Yes, you're right!

Have you read the paper? Do the diagrams make sense to you? (And if so then can you explain please since I feel a bit )

In the paper vertical velocity, when there is an angle (phi) across the y-axis (so up/downhill), is stated as

[math]v_{f,vertical} = v_f \cdot \cos(\beta) \cdot \sin(\phi)[/math]
This makes sense, because

[math]v_{f,x} = v_f \cdot \cos(\beta)[/math]
Because v_f,vertical is perpendicular to the y-axis, we can use v_f,y and sin(phi) to get the opposite side of the side-view triangle, v_f,vertical.

However, I am struggling with defining v_f,vertical when there is an angle (theta) across the slope. Apparently, the correct answer is

[math]v_{f, vertical} = v_f \cdot \sin(\beta) \cdot \sin(\theta)[/math]
But I am not sure why this is the case. I have difficulty with visualising the view that shows this. Maybe you can shine your light on it?!

 

[Steven G]

Line AB is the perpendicular line of the two parallel line with slope theta. This means that the height of point A is the same as the height of point B.

Can someone explain to me how points A & B both have the same height? Are we using the conventual x-y plane?

 

[Cubist]

Can someone explain to me how points A & B both have the same height? Are we using the conventual x-y plane?

It's a 3d situation. All the points have a z coordinate that comes out of the page. A and B having the "same height" means their z coordinates match (I think)

--

However, I'm struggling to just understand Fig 8 (in post#6) which I think is a pre requisite to understanding the figure in post#1. I think this might be a slightly improved version of Fig 8...


I think that the line of steepest slope (of the ground) is in the y axis direction - the y axis direction is shown in Fig8 (b). The angle of the slope is phi.
I also think that the full magnitude of [imath]v_f[/imath] is shown in Fig8(a) but in Fig8(b) this vector is coming "out of the page" a bit.

I'm not sure that I have enough time/ patience to try to work out more - IMO the author(s) of the paper should have spent some extra time working on the diagrams to make this paper so much easier to understand.

In the paper vertical velocity, when there is an angle (phi) across the y-axis (so up/downhill), is stated as

[math]v_{f,vertical} = v_f \cdot \cos(\beta) \cdot \sin(\phi)[/math]
This makes sense, because

[math]v_{f,x} = v_f \cdot \cos(\beta)[/math]

I think this ought to be "y" not "x"?

[math] v_{f,y} = v_f \cdot \cos(\beta) [/math]
EDIT:

However, I am struggling with defining v_f,vertical when there is an angle (theta) across the slope. Apparently, the correct answer is

[math]v_{f, vertical} = v_f \cdot \sin(\beta) \cdot \sin(\theta)[/math]
But I am not sure why this is the case. I have difficulty with visualising the view that shows this. Maybe you can shine your light on it?!

Actually I can see why this is case, and it confirms my thoughts about the slope of the ground (see above). Just imagine a "proper" side view of Fig8(b), one that contains both the y and z axes. (Fig8a isn't actually a side-view, it's a cross sectional view in the direction that the ball is moving).

 

[HJAM24]

I think that the line of steepest slope (of the ground) is in the y axis direction - the y axis direction is shown in Fig8 (b). The angle of the slope is phi.

Correct, the author says the following in the paper:
"Taking the surface of the green to be sloped at angles, with respect to the horizontal, of θ along the x-axis and ϕ along the y-axis"

I think this ought to be "y" not "x"?

Indeed, sorry about that typo.

Just imagine a "proper" side view of Fig8(b), one that contains both the y and z axes

I tried this but when the slope is across the x-axis, the side view will show the slope resulting from θ as a straight vertical line right?

Nevertheless, I would like to thank you very much for your time and help so far. It helped me a lot.

 

[Cubist]

I tried this but when the slope is across the x-axis, the side view will show the slope resulting from θ as a straight vertical line right?

I'm pretty sure that θ is zero, see my reasoning in the next para (I added bold highlights myself to make my thoughts stand out). Although, it really isn't clear. Definitely needs more diagrams IMO

Just after equation (23) the paper says, "The case of an uphill putt on a green of slope ϕ". And earlier in the paper, an uphill putt is defined as... "2.3. Uphill and downhill putts... In the case of straight uphill and downhill putts, θ is equal to zero"

Nevertheless, I would like to thank you very much for your time and help so far. It helped me a lot.

You're welcome. When you win the open please remember the website that helped you with your understanding

 

[Deleted member 4993]

you win the open please remember

Which one US/Australian/GrandSlam..............

 

[Cubist]

Which one US/Australian/GrandSlam..............

I know nothing about golf ?️‍?‍ except occasionally I play crazy golf ?

 

[Cubist]

The OP sent a private message to me, but I'll reply here. Their post...

The project is both for my "green-reading skills" and my Python skills. And a little geometry and physics skills of course .
...
Earlier in the post you mentioned the following regarding the definition of the vertical velocity when putting across a slope: "Actually I can see why this is case". Because you already see it, can you help me see it?

OP, I recommend that you practice drawing the same scene from different, orthogonal, angles. Like this...



AB represents the ball's velocity. Points B and D appear in the same location in the side view but in the plan view you can see that they are actually in different "x" locations. Can you now see how to calculate length DC, and can you see that this length represents the vertical component of the ball's velocity (look at the right angled triangles ABD and ADC).