What is the rate of change of angle for a knight jumping on a chess board?

In summary, the conversation discusses a math challenge problem involving a knight jumping on a chess board and the rate of change of its angle as a function of time relative to an observer. The problem assumes the knight makes a legal move and is solely under the influence of gravity after launch. The goal is to find functions for the angle in the plane of the board and the angle of ascension, as well as the time rates of change for these angles. The problem provides access to an aTan2 function and suggests using calculus to solve it. The participants also discuss the importance of knowing the distance the knight travels and its launch angle in solving the problem.
  • #1
FireStorm000
169
0
Some of my friends, upon finding out that I'd joined these forums, decided to give me a math challenge problem they thought couldn't be solved. Does everyone get the same answer? It goes as follows:

A knight jumps on a chess board (assume it makes a legal move) from point A to point B. Relative to an observing point O, what is the rate of change of angle of the chess piece as a function of time?

Let/Assume:
The piece makes a legal move
The piece is a knight
s (given) : the scale of the chess board. One space has width and height S
C = current position of the piece as a function of time
P = the current position of the piece projected down onto the plane of the chess board
R is a reference point which is always directly right of the center point
[itex]\theta[/itex] (given) is the launch angle
g (constant) is acceleration due to gravity
Coordinate system is defined as the chess board itself, with the origin in one of the corners.
Let theta be the launch angle of the piece.
After launch the piece is solely under the influence of gravity
Find:
A function for the measure of the angle AOP, n (angle in the plane of the board)
A function for the measure of the angle POC, a (accention)
Time rate of change of n, n'
Time rate of change of a, a'
You have access to the aTan2 function, which properly returns the angle between the reference (+X axis) and any point giving an ordered pair IE: aTan2(-2,-1) => 210degrees
Strategy(with calculus)
-I)describe the distance between points A and B, in the plane of the board(d)
-II)derive a function for the vertical component of velocity of the piece, a function of theta, launch velocity, and time
-III)Integrate said function to get a h as a function of time, launch velocity, and launch angle
-IV)Use the definition of velocity to generate a function for s of t and d
-V)Solve the system of eqs from III and IV for initial velocity
-VI)Define the magnitude ||OA|| and ||OB||, rI and rF
-VII)Define the angle between the reference(+X axis) , and OA, OB
-VIII)Write R(t,RI,RF), giving the length of OP at any time
-IX)Take the derivative dR/dt
-X)Realize n is given by ROA plus angle AOP; AOP is determined by creating a right triangle with AO and the component AP perpendicular to AO, and adding their associated rate of change times time.
-XI)Take the derivative with respect to time of n to get n', and one of the answers. Remember your chain rule!
-XII)go back to IX, and realize that a (COP) can be determined from the triangle COP. h = ||CP||, r = ||OP||.
-XIII)once you have a, take the derivative to find the time rate of change of a, a'
I
-I)describe the distance between points A and B, in the plane of the board(d)
knights move up 1 over 2, or up 2 over 1, either way, the distance between A and B is:
d = S*[itex]\sqrt{1^2+2^2}[/itex]
d = S[itex]\sqrt{5}[/itex]
II
-II)derive a function for the vertical component of velocity of the piece, a function of theta, launch velocity, and time
accel=g
Vh=h'=gt+C=gt+V0Sin([itex]\theta[/itex])
III
-III)Integrate said function to get a h as a function of time, launch velocity, and launch angle
h=gt2+V0Sin([itex]\theta[/itex])t+0
(assume time zero is launch)
IV
-IV)Use the definition of velocity to generate a function for s of t and d
distance=rate*time
d/([itex]\Delta[/itex]t) = cos([itex]\theta[/itex])V0
s= cos([itex]\theta[/itex])V0[itex]\Delta[/itex]t
V
-V)Solve the system of equations from III and IV for initial velocity
h=gt2+V0Sin([itex]\theta[/itex])t
0=gt+V0Sin([itex]\theta[/itex])
-gV0 csc([itex]\theta[/itex]) = [itex]\Delta[/itex]t
...
I'll finish the rest of this later...
Answer
View attachment challengeProblem.zip
 
Last edited:
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  • #2
FireStorm000 said:
A knight jumps on a chess board (assume it makes a legal move) from point A to point B. Relative to an observing point O, what is the rate of change of angle of the chess piece as a function of time?
Seems to me it could be anything you want. Nothing in the problem statement tells us how fast the knight moves. It could take a millisecond to get from A to B, or a century.

What am I missing?
 
  • #3
pmsrw3: I think reason why you think you are missing something is because there is no hard coded distance...every square is S x S ; other than that, it is just a projectile shooting from A to B...isn't it?
 
  • #4
I knew I forgot something when I posted. Assume you have the width of a tile, launch angle, and acceleration of gravity, as well as the coordinates of O, A, and B. This is indeed a projectile motion problem.
 
  • #5
:

The rate of change of angle for a knight jumping on a chess board can be determined using the following strategy with calculus:

1. Define the coordinate system: Let the coordinate system be defined as the chess board itself, with the origin in one of the corners. This will allow us to easily track the position of the knight on the board.

2. Define variables: Let s be the scale of the chess board, with one space having a width and height of s. Let C(t) be the current position of the knight as a function of time, and let P(t) be the current position of the knight projected down onto the plane of the chess board. Let R be a reference point that is always directly right of the center point. Let θ be the launch angle of the knight, and let g be the acceleration due to gravity.

3. Determine the angle AOP: The angle AOP is the angle formed between the current position of the knight (C(t)), the origin (O), and the reference point (R). This can be calculated using the aTan2 function, which properly returns the angle between the reference (+X axis) and any point given an ordered pair. Therefore, the function for the measure of angle AOP can be written as: A(t) = aTan2(C(t) - O, R - O).

4. Determine the angle POC: The angle POC is the angle formed between the current position of the knight (C(t)), the origin (O), and the projection point (P(t)). This can also be calculated using the aTan2 function, with the ordered pair being (C(t) - O, P(t) - O). Therefore, the function for the measure of angle POC can be written as: a(t) = aTan2(C(t) - O, P(t) - O).

5. Determine the time rate of change of AOP: To find the time rate of change of AOP, we can use the chain rule of calculus. The chain rule states that if y = f(g(x)), then y' = f'(g(x)) * g'(x). In this case, A(t) is a function of C(t), so the time rate of change of AOP can be calculated as: A'(t) = aTan2'(C(t) - O, R - O) * C'(t).

6. Determine the time rate of change
 

1. What is angular velocity and how is it calculated?

Angular velocity is a measure of how fast an object is rotating around a fixed axis. It is calculated by dividing the change in angular displacement by the change in time. The formula for angular velocity is ω = Δθ/Δt, where ω is the angular velocity, Δθ is the change in angular displacement, and Δt is the change in time.

2. What is the relationship between angular velocity and linear velocity?

Angular velocity and linear velocity are related by the radius of the circle or arc that the object is moving along. The linear velocity is equal to the angular velocity multiplied by the radius. The formula for this relationship is v = ωr, where v is the linear velocity, ω is the angular velocity, and r is the radius.

3. How is kinematics used to solve problems involving angular velocity?

Kinematics is the study of motion, and it can be used to solve problems involving angular velocity by applying the same principles used to solve problems involving linear motion. This includes using equations such as ω = Δθ/Δt, v = ωr, and a = ω/t, where ω is the angular velocity, Δθ is the change in angular displacement, Δt is the change in time, v is the linear velocity, r is the radius, and a is the angular acceleration.

4. Can angular velocity be negative?

Yes, angular velocity can be negative. A negative angular velocity indicates that the object is rotating in the opposite direction as a positive angular velocity. For example, a counterclockwise rotation would have a positive angular velocity, while a clockwise rotation would have a negative angular velocity.

5. How is calculus used in solving problems involving angular velocity?

Calculus is used to solve problems involving angular velocity by finding the derivative of the angular displacement with respect to time. This derivative, known as angular velocity, can then be used to find other quantities such as linear velocity and angular acceleration. Calculus is also used to find the area under a curve, which can be used to determine the total angular displacement of an object.

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