Mixing
Help calculating a point in 2d space on a rotating right angle triangle
$begingroup$
I am trying to solve for a point on a triangle that rotates around a fixed point. For ease of calculation this point is at X: 0 Y: 0
The length and width of the triangle are always fixed, the only thing that changes is the rotation angle. How can I solve for X: ??? and Y: ??? taking into account the rotation angle?
I am trying to place a 2d rendered object in Java so I need to break the formula down to it's base elements so I can put it into code and have the formula spit out X and Y coordinates
The known variables are all listed in the pictures. The angle of the triangle, the length & width. Anything else that is needed to find the ??? must be calculated first I guess.
Please don't just link wiki pages. I have read them. I don't understand them. Hence why I am asking for help walking through the solution.
Thank you!
geometry trigonometry
edited Dec 21 '15 at 15:57
Community♦
1
asked Dec 21 '15 at 13:48
Gummby8Gummby8
61
$endgroup$
add a comment |
$begingroup$
I am trying to solve for a point on a triangle that rotates around a fixed point. For ease of calculation this point is at X: 0 Y: 0
The length and width of the triangle are always fixed, the only thing that changes is the rotation angle. How can I solve for X: ??? and Y: ??? taking into account the rotation angle?
I am trying to place a 2d rendered object in Java so I need to break the formula down to it's base elements so I can put it into code and have the formula spit out X and Y coordinates
The known variables are all listed in the pictures. The angle of the triangle, the length & width. Anything else that is needed to find the ??? must be calculated first I guess.
Please don't just link wiki pages. I have read them. I don't understand them. Hence why I am asking for help walking through the solution.
Thank you!
geometry trigonometry
edited Dec 21 '15 at 15:57
Community♦
1
asked Dec 21 '15 at 13:48
Gummby8Gummby8
61
$endgroup$
$begingroup$
Which parameters do you want to give your triangle? For example, do you want to give it the current angle, two sides and the enclosed angle between those two sides?
$endgroup$
– Imago
Dec 21 '15 at 13:58
$begingroup$
Multiply the coordinates with a rotation matrix en.m.wikipedia.org/wiki/Rotation_matrix
$endgroup$
– David Quinn
Dec 21 '15 at 14:25
add a comment |
$begingroup$
I am trying to solve for a point on a triangle that rotates around a fixed point. For ease of calculation this point is at X: 0 Y: 0
The length and width of the triangle are always fixed, the only thing that changes is the rotation angle. How can I solve for X: ??? and Y: ??? taking into account the rotation angle?
I am trying to place a 2d rendered object in Java so I need to break the formula down to it's base elements so I can put it into code and have the formula spit out X and Y coordinates
The known variables are all listed in the pictures. The angle of the triangle, the length & width. Anything else that is needed to find the ??? must be calculated first I guess.
Please don't just link wiki pages. I have read them. I don't understand them. Hence why I am asking for help walking through the solution.
Thank you!
geometry trigonometry
edited Dec 21 '15 at 15:57
Community♦
1
asked Dec 21 '15 at 13:48
Gummby8Gummby8
61
$endgroup$
I am trying to solve for a point on a triangle that rotates around a fixed point. For ease of calculation this point is at X: 0 Y: 0
The length and width of the triangle are always fixed, the only thing that changes is the rotation angle. How can I solve for X: ??? and Y: ??? taking into account the rotation angle?
I am trying to place a 2d rendered object in Java so I need to break the formula down to it's base elements so I can put it into code and have the formula spit out X and Y coordinates
The known variables are all listed in the pictures. The angle of the triangle, the length & width. Anything else that is needed to find the ??? must be calculated first I guess.
Please don't just link wiki pages. I have read them. I don't understand them. Hence why I am asking for help walking through the solution.
Thank you!
geometry trigonometry
geometry trigonometry
edited Dec 21 '15 at 15:57
Community♦
1
asked Dec 21 '15 at 13:48
Gummby8Gummby8
61
edited Dec 21 '15 at 15:57
Community♦
1
asked Dec 21 '15 at 13:48
Gummby8Gummby8
61
edited Dec 21 '15 at 15:57
Community♦
1
edited Dec 21 '15 at 15:57
Community♦
1
edited Dec 21 '15 at 15:57
Community♦
1
1
asked Dec 21 '15 at 13:48
Gummby8Gummby8
61
asked Dec 21 '15 at 13:48
Gummby8Gummby8
61
asked Dec 21 '15 at 13:48
Gummby8Gummby8
61
61
$begingroup$
Which parameters do you want to give your triangle? For example, do you want to give it the current angle, two sides and the enclosed angle between those two sides?
$endgroup$
– Imago
Dec 21 '15 at 13:58
$begingroup$
Multiply the coordinates with a rotation matrix en.m.wikipedia.org/wiki/Rotation_matrix
$endgroup$
– David Quinn
Dec 21 '15 at 14:25
add a comment |
$begingroup$
Which parameters do you want to give your triangle? For example, do you want to give it the current angle, two sides and the enclosed angle between those two sides?
$endgroup$
– Imago
Dec 21 '15 at 13:58
$begingroup$
Multiply the coordinates with a rotation matrix en.m.wikipedia.org/wiki/Rotation_matrix
$endgroup$
– David Quinn
Dec 21 '15 at 14:25
$begingroup$
Which parameters do you want to give your triangle? For example, do you want to give it the current angle, two sides and the enclosed angle between those two sides?
$endgroup$
– Imago
Dec 21 '15 at 13:58
$begingroup$
Which parameters do you want to give your triangle? For example, do you want to give it the current angle, two sides and the enclosed angle between those two sides?
$endgroup$
– Imago
Dec 21 '15 at 13:58
$begingroup$
Multiply the coordinates with a rotation matrix en.m.wikipedia.org/wiki/Rotation_matrix
$endgroup$
– David Quinn
Dec 21 '15 at 14:25
$begingroup$
Multiply the coordinates with a rotation matrix en.m.wikipedia.org/wiki/Rotation_matrix
$endgroup$
– David Quinn
Dec 21 '15 at 14:25
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Imagine the hypotenuse of your triangle is the radius of a circle for which the length can be calculated with the Pythagorean theorem:
$$
H = sqrt{L^2 + W^2} \
$$
Then, instead of calculating the angle between the negative $y$-axis and the hypotenuse as drawn, imagine the initial angle of your hypotenuse is the angle between the $x$-axis and the hypotenuse which can be calculated with the $tan$ function and its inverse $arctan$ function:
$$
tan theta = frac{textrm{opposite}}{textrm{adjacent}} = frac{L}{W} \
theta = arctan frac{L}{W} \
$$
Finally, the coordinates can be calculated by multiplying the hypotenuse ($H$) with the appropriate trigonometric function ($sin$ or $cos$) applied to the complete angle ($theta + Angle$) which can then be simplified:
$$
X = H cos(theta + Angle) = W cos(Angle) - L sin(Angle) \
Y = H sin(theta + Angle) = L cos(Angle) + W sin(Angle) \
$$
Let's try these equations with the values from your first triangle where $Angle$ is $0$:
$$
X = 3 cos(0) - (-5) sin(0) = 3 \
Y = (-5) cos(0) + 3 sin(0) = -5 \
$$
Now let's try again with the values from your second triangle where $Angle$ is $35^circ$ converted to radians:
$$
X = 3 cos(frac{35 pi}{180}) - (-5) sin(frac{35 pi}{180}) approx 5.33 \
Y = (-5) cos(frac{35 pi}{180}) + 3 sin(frac{35 pi}{180}) approx -2.38 \
$$
answered Dec 21 '15 at 23:39
cr3cr3
1,345511
$endgroup$
add a comment |
$begingroup$
Cr3 is correct but I found a bit smaller way to do it from a different help forums. Thank you all
x′= Width * cos(Angle) - Length * sin(Angle)
y′= Width * sin(Angle) + Length * cos(Angle)
x′=3cos(35)+5sin(35)≈5.33
y′=3sin(35)−5cos(35)≈−2.38
answered Dec 22 '15 at 17:18
Gummby8Gummby8
31
$endgroup$
$begingroup$
If it looks like your answer is a bit smaller because it uses degrees ($35^circ$) whereas my answer uses radians ($frac{35pi}{180}$), please note that theMath.cosfunction in Java expects an angle parameter in radians.
$endgroup$
– cr3
Dec 23 '15 at 19:19
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Imagine the hypotenuse of your triangle is the radius of a circle for which the length can be calculated with the Pythagorean theorem:
$$
H = sqrt{L^2 + W^2} \
$$
Then, instead of calculating the angle between the negative $y$-axis and the hypotenuse as drawn, imagine the initial angle of your hypotenuse is the angle between the $x$-axis and the hypotenuse which can be calculated with the $tan$ function and its inverse $arctan$ function:
$$
tan theta = frac{textrm{opposite}}{textrm{adjacent}} = frac{L}{W} \
theta = arctan frac{L}{W} \
$$
Finally, the coordinates can be calculated by multiplying the hypotenuse ($H$) with the appropriate trigonometric function ($sin$ or $cos$) applied to the complete angle ($theta + Angle$) which can then be simplified:
$$
X = H cos(theta + Angle) = W cos(Angle) - L sin(Angle) \
Y = H sin(theta + Angle) = L cos(Angle) + W sin(Angle) \
$$
Let's try these equations with the values from your first triangle where $Angle$ is $0$:
$$
X = 3 cos(0) - (-5) sin(0) = 3 \
Y = (-5) cos(0) + 3 sin(0) = -5 \
$$
Now let's try again with the values from your second triangle where $Angle$ is $35^circ$ converted to radians:
$$
X = 3 cos(frac{35 pi}{180}) - (-5) sin(frac{35 pi}{180}) approx 5.33 \
Y = (-5) cos(frac{35 pi}{180}) + 3 sin(frac{35 pi}{180}) approx -2.38 \
$$
answered Dec 21 '15 at 23:39
cr3cr3
1,345511
$endgroup$
add a comment |
$begingroup$
Imagine the hypotenuse of your triangle is the radius of a circle for which the length can be calculated with the Pythagorean theorem:
$$
H = sqrt{L^2 + W^2} \
$$
Then, instead of calculating the angle between the negative $y$-axis and the hypotenuse as drawn, imagine the initial angle of your hypotenuse is the angle between the $x$-axis and the hypotenuse which can be calculated with the $tan$ function and its inverse $arctan$ function:
$$
tan theta = frac{textrm{opposite}}{textrm{adjacent}} = frac{L}{W} \
theta = arctan frac{L}{W} \
$$
Finally, the coordinates can be calculated by multiplying the hypotenuse ($H$) with the appropriate trigonometric function ($sin$ or $cos$) applied to the complete angle ($theta + Angle$) which can then be simplified:
$$
X = H cos(theta + Angle) = W cos(Angle) - L sin(Angle) \
Y = H sin(theta + Angle) = L cos(Angle) + W sin(Angle) \
$$
Let's try these equations with the values from your first triangle where $Angle$ is $0$:
$$
X = 3 cos(0) - (-5) sin(0) = 3 \
Y = (-5) cos(0) + 3 sin(0) = -5 \
$$
Now let's try again with the values from your second triangle where $Angle$ is $35^circ$ converted to radians:
$$
X = 3 cos(frac{35 pi}{180}) - (-5) sin(frac{35 pi}{180}) approx 5.33 \
Y = (-5) cos(frac{35 pi}{180}) + 3 sin(frac{35 pi}{180}) approx -2.38 \
$$
answered Dec 21 '15 at 23:39
cr3cr3
1,345511
$endgroup$
add a comment |
$begingroup$
Imagine the hypotenuse of your triangle is the radius of a circle for which the length can be calculated with the Pythagorean theorem:
$$
H = sqrt{L^2 + W^2} \
$$
Then, instead of calculating the angle between the negative $y$-axis and the hypotenuse as drawn, imagine the initial angle of your hypotenuse is the angle between the $x$-axis and the hypotenuse which can be calculated with the $tan$ function and its inverse $arctan$ function:
$$
tan theta = frac{textrm{opposite}}{textrm{adjacent}} = frac{L}{W} \
theta = arctan frac{L}{W} \
$$
Finally, the coordinates can be calculated by multiplying the hypotenuse ($H$) with the appropriate trigonometric function ($sin$ or $cos$) applied to the complete angle ($theta + Angle$) which can then be simplified:
$$
X = H cos(theta + Angle) = W cos(Angle) - L sin(Angle) \
Y = H sin(theta + Angle) = L cos(Angle) + W sin(Angle) \
$$
Let's try these equations with the values from your first triangle where $Angle$ is $0$:
$$
X = 3 cos(0) - (-5) sin(0) = 3 \
Y = (-5) cos(0) + 3 sin(0) = -5 \
$$
Now let's try again with the values from your second triangle where $Angle$ is $35^circ$ converted to radians:
$$
X = 3 cos(frac{35 pi}{180}) - (-5) sin(frac{35 pi}{180}) approx 5.33 \
Y = (-5) cos(frac{35 pi}{180}) + 3 sin(frac{35 pi}{180}) approx -2.38 \
$$
answered Dec 21 '15 at 23:39
cr3cr3
1,345511
$endgroup$
Imagine the hypotenuse of your triangle is the radius of a circle for which the length can be calculated with the Pythagorean theorem:
$$
H = sqrt{L^2 + W^2} \
$$
Then, instead of calculating the angle between the negative $y$-axis and the hypotenuse as drawn, imagine the initial angle of your hypotenuse is the angle between the $x$-axis and the hypotenuse which can be calculated with the $tan$ function and its inverse $arctan$ function:
$$
tan theta = frac{textrm{opposite}}{textrm{adjacent}} = frac{L}{W} \
theta = arctan frac{L}{W} \
$$
Finally, the coordinates can be calculated by multiplying the hypotenuse ($H$) with the appropriate trigonometric function ($sin$ or $cos$) applied to the complete angle ($theta + Angle$) which can then be simplified:
$$
X = H cos(theta + Angle) = W cos(Angle) - L sin(Angle) \
Y = H sin(theta + Angle) = L cos(Angle) + W sin(Angle) \
$$
Let's try these equations with the values from your first triangle where $Angle$ is $0$:
$$
X = 3 cos(0) - (-5) sin(0) = 3 \
Y = (-5) cos(0) + 3 sin(0) = -5 \
$$
Now let's try again with the values from your second triangle where $Angle$ is $35^circ$ converted to radians:
$$
X = 3 cos(frac{35 pi}{180}) - (-5) sin(frac{35 pi}{180}) approx 5.33 \
Y = (-5) cos(frac{35 pi}{180}) + 3 sin(frac{35 pi}{180}) approx -2.38 \
$$
answered Dec 21 '15 at 23:39
cr3cr3
1,345511
edited Dec 22 '15 at 0:16
answered Dec 21 '15 at 23:39
cr3cr3
1,345511
answered Dec 21 '15 at 23:39
cr3cr3
1,345511
answered Dec 21 '15 at 23:39
cr3cr3
1,345511
1,345511
add a comment |
add a comment |
$begingroup$
Cr3 is correct but I found a bit smaller way to do it from a different help forums. Thank you all
x′= Width * cos(Angle) - Length * sin(Angle)
y′= Width * sin(Angle) + Length * cos(Angle)
x′=3cos(35)+5sin(35)≈5.33
y′=3sin(35)−5cos(35)≈−2.38
answered Dec 22 '15 at 17:18
Gummby8Gummby8
31
$endgroup$
$begingroup$
If it looks like your answer is a bit smaller because it uses degrees ($35^circ$) whereas my answer uses radians ($frac{35pi}{180}$), please note that theMath.cosfunction in Java expects an angle parameter in radians.
$endgroup$
– cr3
Dec 23 '15 at 19:19
add a comment |
$begingroup$
Cr3 is correct but I found a bit smaller way to do it from a different help forums. Thank you all
x′= Width * cos(Angle) - Length * sin(Angle)
y′= Width * sin(Angle) + Length * cos(Angle)
x′=3cos(35)+5sin(35)≈5.33
y′=3sin(35)−5cos(35)≈−2.38
answered Dec 22 '15 at 17:18
Gummby8Gummby8
31
$endgroup$
$begingroup$
If it looks like your answer is a bit smaller because it uses degrees ($35^circ$) whereas my answer uses radians ($frac{35pi}{180}$), please note that theMath.cosfunction in Java expects an angle parameter in radians.
$endgroup$
– cr3
Dec 23 '15 at 19:19
add a comment |
$begingroup$
Cr3 is correct but I found a bit smaller way to do it from a different help forums. Thank you all
x′= Width * cos(Angle) - Length * sin(Angle)
y′= Width * sin(Angle) + Length * cos(Angle)
x′=3cos(35)+5sin(35)≈5.33
y′=3sin(35)−5cos(35)≈−2.38
answered Dec 22 '15 at 17:18
Gummby8Gummby8
31
$endgroup$
Cr3 is correct but I found a bit smaller way to do it from a different help forums. Thank you all
x′= Width * cos(Angle) - Length * sin(Angle)
y′= Width * sin(Angle) + Length * cos(Angle)
x′=3cos(35)+5sin(35)≈5.33
y′=3sin(35)−5cos(35)≈−2.38
answered Dec 22 '15 at 17:18
Gummby8Gummby8
31
answered Dec 22 '15 at 17:18
Gummby8Gummby8
31
answered Dec 22 '15 at 17:18
Gummby8Gummby8
31
answered Dec 22 '15 at 17:18
Gummby8Gummby8
31
31
$begingroup$
If it looks like your answer is a bit smaller because it uses degrees ($35^circ$) whereas my answer uses radians ($frac{35pi}{180}$), please note that theMath.cosfunction in Java expects an angle parameter in radians.
$endgroup$
– cr3
Dec 23 '15 at 19:19
add a comment |
$begingroup$
If it looks like your answer is a bit smaller because it uses degrees ($35^circ$) whereas my answer uses radians ($frac{35pi}{180}$), please note that theMath.cosfunction in Java expects an angle parameter in radians.
$endgroup$
– cr3
Dec 23 '15 at 19:19
$begingroup$
If it looks like your answer is a bit smaller because it uses degrees ($35^circ$) whereas my answer uses radians ($frac{35pi}{180}$), please note that the
Math.cos function in Java expects an angle parameter in radians.$endgroup$
– cr3
Dec 23 '15 at 19:19
$begingroup$
If it looks like your answer is a bit smaller because it uses degrees ($35^circ$) whereas my answer uses radians ($frac{35pi}{180}$), please note that the
Math.cos function in Java expects an angle parameter in radians.$endgroup$
– cr3
Dec 23 '15 at 19:19
add a comment |
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$begingroup$
Which parameters do you want to give your triangle? For example, do you want to give it the current angle, two sides and the enclosed angle between those two sides?
$endgroup$
– Imago
Dec 21 '15 at 13:58
$begingroup$
Multiply the coordinates with a rotation matrix en.m.wikipedia.org/wiki/Rotation_matrix
$endgroup$
– David Quinn
Dec 21 '15 at 14:25