Geometry... find a(x)=?
I have fixed h and r. The red string is pulled to the right away from point 0. The rope is wrapped around a drum thus causing it to rotate. At about x=3, the drum has been rotated by a-radians. For clarity, the color of the same string is changed to green. Please see the attached image.
Question: Can you find function a(x) in terms of constants h and r?
Thank you
Cam-1.png
Question: Can you find function a(x) in terms of constants h and r?
Thank you
Cam-1.png
ASKER CERTIFIED SOLUTION
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ASKER
The red string increases in height by k.
k=k = r*sin(b)
Also, the horzontal base of our right-triangle in creases by m.
m = r-(r*cos(b))
The hypotenuse increases by L.
L = b*r
assuming b is in radians.
k=k = r*sin(b)
Also, the horzontal base of our right-triangle in creases by m.
m = r-(r*cos(b))
The hypotenuse increases by L.
L = b*r
assuming b is in radians.
ASKER
My objective is to write a program in java or C++ to supply x then get it ploted and have angle a as an output in degrees or radians. I am writting this because your are strong in C++ and I am new to java.
Do you think you have the time to help me in this exercise? If so, please let me knowwhere I should be posting? Of course after the geometry relations are handled here to your satisfaction.
Thank you.
Do you think you have the time to help me in this exercise? If so, please let me knowwhere I should be posting? Of course after the geometry relations are handled here to your satisfaction.
Thank you.
ASKER
also, at some point it will be very nice if we could assign the rate of x-pull (like 4 inch/sec) and watch it animate.
Hi,
In my previous post, the excess rope I referred to was actually the case if there was a spindle at the right-most point of the circle. Sorry about that. I believe the problem needs a few more relationships than the ones you wrote in order to arrive at angle 'a' after we figure out angle 'b'. I realized that after you showed your second diagram.
>> (h+L)^2 = (h + k)^2 + (x + m)^2
This appears to be a small angle (b) approximation, so that L, the length of the arc subscribed by 'b', becomes the hypontenuse of the triangle in the circle (not shown).
From your goals, you want x to grow resulting in larger angle 'b'. (In fact x can keep growing way past C = 2 pi r (assuming enough rope is looped around the barrel). So, when I sketched a diagram where 'b' was about 60 degrees (pi/3 radians), I think that small angle approximation relationship is inaccurate.
If this is in line with your thinking, then I can suggest some notation for another diagram and we can work from there since there are a lot of equations to work with.
In my previous post, the excess rope I referred to was actually the case if there was a spindle at the right-most point of the circle. Sorry about that. I believe the problem needs a few more relationships than the ones you wrote in order to arrive at angle 'a' after we figure out angle 'b'. I realized that after you showed your second diagram.
>> (h+L)^2 = (h + k)^2 + (x + m)^2
This appears to be a small angle (b) approximation, so that L, the length of the arc subscribed by 'b', becomes the hypontenuse of the triangle in the circle (not shown).
From your goals, you want x to grow resulting in larger angle 'b'. (In fact x can keep growing way past C = 2 pi r (assuming enough rope is looped around the barrel). So, when I sketched a diagram where 'b' was about 60 degrees (pi/3 radians), I think that small angle approximation relationship is inaccurate.
If this is in line with your thinking, then I can suggest some notation for another diagram and we can work from there since there are a lot of equations to work with.
ASKER
re:> needs a few more relationships than the ones you wrote in order to arrive at angle 'a' ...
You are right. In my next post I will add some addition I have.
re:> small angle (b) approximation...
There no approximation. This is the definition of radian: Arch length = radius * angle in radian
re:> you want x to grow resulting in larger angle 'b'
Because the puuled string has to be tangent to the drum, it possible cannot go beyond 75 degrees even if X was very large. In our case is X will be max 45 inches thus possible reaching no more than 30 degrees (even if drum is small and has to rotate more than 2rPI.
brb
You are right. In my next post I will add some addition I have.
re:> small angle (b) approximation...
There no approximation. This is the definition of radian: Arch length = radius * angle in radian
re:> you want x to grow resulting in larger angle 'b'
Because the puuled string has to be tangent to the drum, it possible cannot go beyond 75 degrees even if X was very large. In our case is X will be max 45 inches thus possible reaching no more than 30 degrees (even if drum is small and has to rotate more than 2rPI.
brb
ASKER
Below the information is for your consideration to see if they can be of any use towards the solution a(x).
Trianle P3-P5-P2 is right angle at p2 (because pulled string is tangent to the circle thus radius p5-p2 is perpendicualr to it).
Another right-angle triangle is p5-p4-p3:
H= from p5 to P3
H^2 = (r+x)^2 + h^2
Also using (h+b*r)^2 from previous posts:
H^2 = r^2 + (h+b*r)^2
or:
(r+x)^2 + h^2 = r^2 + (h+b*r)^2
r^2 + x^2 + 2rx = r^2 + h^2 + (br)^2 + 2hbr
x^2 + 2rx = h^2 + (br)^2 + 2hbr
---------------------
What we know about angle a is tha if we were to pull the string dwon. say y distance, then y-length unwrapped divided by r gives angle a in radians:
a in radian = y/r
but it is not this simple. if the tangent point was able to remain at its intial location (not possible by definition) then lenght of the pulled string minus h could be divided by rto give angle a in radians.
However, the tangent point is moving. The movement of the tangent point (where the string and the drum are first touching) has to be related t some other dimentions like x, r, and h in order to fine angle of rotation a.
Trianle P3-P5-P2 is right angle at p2 (because pulled string is tangent to the circle thus radius p5-p2 is perpendicualr to it).
Another right-angle triangle is p5-p4-p3:
H= from p5 to P3
H^2 = (r+x)^2 + h^2
Also using (h+b*r)^2 from previous posts:
H^2 = r^2 + (h+b*r)^2
or:
(r+x)^2 + h^2 = r^2 + (h+b*r)^2
r^2 + x^2 + 2rx = r^2 + h^2 + (br)^2 + 2hbr
x^2 + 2rx = h^2 + (br)^2 + 2hbr
---------------------
What we know about angle a is tha if we were to pull the string dwon. say y distance, then y-length unwrapped divided by r gives angle a in radians:
a in radian = y/r
but it is not this simple. if the tangent point was able to remain at its intial location (not possible by definition) then lenght of the pulled string minus h could be divided by rto give angle a in radians.
However, the tangent point is moving. The movement of the tangent point (where the string and the drum are first touching) has to be related t some other dimentions like x, r, and h in order to fine angle of rotation a.
ASKER
see the graph:
Cam-3.png
Cam-3.png
ASKER
sorry lastone was wrong one...
Cam-3.png
Cam-3.png
ASKER
x^2 + 2rx = h^2 + (br)^2 + 2hbr
x^2 + 2rx r^2= r^2+ h^2 + (br)^2 + 2hbr
(x + r)^2 = r^2(1 + b^2) + h(h + 2br)
x^2 + 2rx r^2= r^2+ h^2 + (br)^2 + 2hbr
(x + r)^2 = r^2(1 + b^2) + h(h + 2br)
ASKER
x^2 + 2rx = h^2 + (br)^2 + 2hbr
x^2 + 2rx + r^2= r^2+ h^2 + (br)^2 + 2hbr
(x + r)^2 = r^2(1 + b^2) + h(h + 2br)
x^2 + 2rx + r^2= r^2+ h^2 + (br)^2 + 2hbr
(x + r)^2 = r^2(1 + b^2) + h(h + 2br)
>Trianle P3-P5-P2 is right angle at p2 (because pulled string is tangent to the circle thus radius p5-p2 is perpendicualr to it).
Where is P5?
Is P1 the point on the drum where the string contacts the drum tangentially?
Is P2 the point on the green line where it crosses the upper blue horizontal line? If not, then please label that point?
Where is P5?
Is P1 the point on the drum where the string contacts the drum tangentially?
Is P2 the point on the green line where it crosses the upper blue horizontal line? If not, then please label that point?
ASKER
String pulled a bit further.
Cam-3a.png
Cam-3a.png
Is P6 the point on the drum where the string contacts the drum tangentially?
Is P5 the axis of drum ?
Please give us label for the point on the green line where it crosses the upper blue horizontal line?
Is P5 the axis of drum ?
Please give us label for the point on the green line where it crosses the upper blue horizontal line?
ASKER
re:> Please give us label for the point on the green line where it crosses the upper blue horizontal line?
There is none for that location but you can asigne one there or anywhere else.
P1: On the drum where red sting seperates from the drum.
P2: On the drum where solid green sting seperates from the drum.
P3: Solid green string on x-axis
P4: Projection of drum center on x=axis.
p5: Center of the drum.
p6: On the drum where dashed green sting seperates from the drum.
P7: Dashed green string on x-axis
There is none for that location but you can asigne one there or anywhere else.
P1: On the drum where red sting seperates from the drum.
P2: On the drum where solid green sting seperates from the drum.
P3: Solid green string on x-axis
P4: Projection of drum center on x=axis.
p5: Center of the drum.
p6: On the drum where dashed green sting seperates from the drum.
P7: Dashed green string on x-axis
ASKER
If the red string is pulled downward only (in y-direction), then angle a is:
a in radians = (pull in y-axis)/r
but x-pulls creates angle b and reduces a-angle rotation.
So, if we can find the p2-p3 length minus h and call it = DL (delta L)
r(a + b) = DL
a in radians = (pull in y-axis)/r
but x-pulls creates angle b and reduces a-angle rotation.
So, if we can find the p2-p3 length minus h and call it = DL (delta L)
r(a + b) = DL
Hi,
I'm leaving for awhile today, so I'll look at your more detailed posts when I get back.
>> L = the length of the arch from p1 to p2 = b*r
agreed (by definition)
>> (h+L)^2 = (h + k)^2 + (x + m)^2
please explain how you got this (it does not seem correct to me).
Why do you say that the max angle 'b' is 75 degrees? Where did 75 come from?
Since you have symbols, h and r, as well as numbers (e.g., for x and max length of rope), please give (typical) values for h and r.
In http:#36281780 , did you notice that angle P6_P5_P1 = complement of angle P6_P7_P4.
I'm leaving for awhile today, so I'll look at your more detailed posts when I get back.
>> L = the length of the arch from p1 to p2 = b*r
agreed (by definition)
>> (h+L)^2 = (h + k)^2 + (x + m)^2
please explain how you got this (it does not seem correct to me).
Why do you say that the max angle 'b' is 75 degrees? Where did 75 come from?
Since you have symbols, h and r, as well as numbers (e.g., for x and max length of rope), please give (typical) values for h and r.
In http:#36281780 , did you notice that angle P6_P5_P1 = complement of angle P6_P7_P4.
ASKER
Simplification:
I have added a pulley (orange in color) to keep seperation point of the spring from the drum at a fixed point. After a our animation program (in c++ or java) is developed, we could take a look at the version with no such pulley.
So now we have triangle ABC where:
h^2 + (BC)^2 = AC^2
AC = h + DL where DL is the length of string added to h to make AC
h^2 + X^2 = (h + DL)^2
SQR(h^2 + X^2) = h + DL
SQR(h^2 + X^2) - h = DL (eg. 1)
Also we have:
DL = r*a (eg. 2)
substituting DL from eg. 2 into eg. 1:
SQR(h^2 + X^2) - h = r*a
a = (SQR(h^2 + X^2) - h)/r
a = (SQR(3600 + X^2))/10
Cam-1a.png
I have added a pulley (orange in color) to keep seperation point of the spring from the drum at a fixed point. After a our animation program (in c++ or java) is developed, we could take a look at the version with no such pulley.
So now we have triangle ABC where:
h^2 + (BC)^2 = AC^2
AC = h + DL where DL is the length of string added to h to make AC
h^2 + X^2 = (h + DL)^2
SQR(h^2 + X^2) = h + DL
SQR(h^2 + X^2) - h = DL (eg. 1)
Also we have:
DL = r*a (eg. 2)
substituting DL from eg. 2 into eg. 1:
SQR(h^2 + X^2) - h = r*a
a = (SQR(h^2 + X^2) - h)/r
a = (SQR(3600 + X^2))/10
Cam-1a.png
I'm back (briefly). I see you have simplified to the assumption made in my first post that you have a spindle. Makes life easier (although I should be able to help with the more complicated problem after we work this one out).
Your equation:
>> a = (SQR(h^2 + X^2) - h)/r
is consistent with my equation in the first post:
>> a(x) = ( sqrt( h^2 + x^2 ) - h ) / r
so, we're on the same page.
>> at some point it will be very nice if we could assign the rate of x-pull (like 4 inch/sec) and watch it animate.
Since you want to work with animation, I would choose Java over C++ to get started quickly. Standard C++, as you know, has no graphics defined, whereas Java does. Probably best C++ graphics comes with OpenGL (but different installations for different platforms), but that's yet another installation and learning experience. Java has platform-independent Swing (and probably more since I worked with it last century); and then it is also easy to convert the application to an applet for your internet browser, and share results with the rest of the world.
I'd suggest a first step of writing a simple java program that generates a tabular listing of a(x) vs x, where x runs from 0 to 62.8 inches (for a full 2 pi r) at increments of .1 or .25. This will help verify that we got the equations right.
I'll be back briefly tomorrow, and will be available better on Monday. It just so happens that I picked up a Java book to refresh what I knew a decade ago, so I should be able to help. I would think that by Monday, we should have this tabular listing worked out. Since you indicate that you are learning Java now, this should be an easy starting point.
If you are interested in angular velocities and accelerations, that can be computed numerically, or analytically using with differentiation.
See you soon.
Your equation:
>> a = (SQR(h^2 + X^2) - h)/r
is consistent with my equation in the first post:
>> a(x) = ( sqrt( h^2 + x^2 ) - h ) / r
so, we're on the same page.
>> at some point it will be very nice if we could assign the rate of x-pull (like 4 inch/sec) and watch it animate.
Since you want to work with animation, I would choose Java over C++ to get started quickly. Standard C++, as you know, has no graphics defined, whereas Java does. Probably best C++ graphics comes with OpenGL (but different installations for different platforms), but that's yet another installation and learning experience. Java has platform-independent Swing (and probably more since I worked with it last century); and then it is also easy to convert the application to an applet for your internet browser, and share results with the rest of the world.
I'd suggest a first step of writing a simple java program that generates a tabular listing of a(x) vs x, where x runs from 0 to 62.8 inches (for a full 2 pi r) at increments of .1 or .25. This will help verify that we got the equations right.
I'll be back briefly tomorrow, and will be available better on Monday. It just so happens that I picked up a Java book to refresh what I knew a decade ago, so I should be able to help. I would think that by Monday, we should have this tabular listing worked out. Since you indicate that you are learning Java now, this should be an easy starting point.
If you are interested in angular velocities and accelerations, that can be computed numerically, or analytically using with differentiation.
See you soon.
ASKER
phoffric,
I love all you have said. It will be nice to include few JText controls for x/sec, h, and r. As you have mentioned, for now a tabular output will be fantastic. Later on we could have it animate at 4 inch/sec you have discussed.
I should have listened to you earlier on the simplified formula. Also, friend of mine apparently has been able to figure out the equation without this simplification. I haven't studied yet. I will run by you in another post after we finish this first phase.
Thank you very much.
Mike
I love all you have said. It will be nice to include few JText controls for x/sec, h, and r. As you have mentioned, for now a tabular output will be fantastic. Later on we could have it animate at 4 inch/sec you have discussed.
I should have listened to you earlier on the simplified formula. Also, friend of mine apparently has been able to figure out the equation without this simplification. I haven't studied yet. I will run by you in another post after we finish this first phase.
Thank you very much.
Mike
SOLUTION
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ASKER
re:> What graphics software did you use to create your images?
MS Paint.
re:> Do these numbers make sense?
The drum is not circular. It is totally different shape. I have simplified enough to get moving and deal with the original shape later. So, I wouldn't expect any sense of accuracy from the numbers produced at this point.
Thank you for your time and effort on this animation.
Mike
MS Paint.
re:> Do these numbers make sense?
The drum is not circular. It is totally different shape. I have simplified enough to get moving and deal with the original shape later. So, I wouldn't expect any sense of accuracy from the numbers produced at this point.
Thank you for your time and effort on this animation.
Mike
>> a tabular listing of a(x) vs x, where x runs from 0 to 62.8 inches
Meant to say, (referencing first post)
where s = lg - h runs from 0 to 62.8 inches and solve (easily) for x to get max value.
I believe that I had a cos and a tan in my equations in the original problem, so the rrz@871311 equations be on track. But that form still needs work in order to produce a constant dx/dt. But, working on the simpler problem will help in understanding the more complicated one.
Have a good night.
Paul
Meant to say, (referencing first post)
where s = lg - h runs from 0 to 62.8 inches and solve (easily) for x to get max value.
I believe that I had a cos and a tan in my equations in the original problem, so the rrz@871311 equations be on track. But that form still needs work in order to produce a constant dx/dt. But, working on the simpler problem will help in understanding the more complicated one.
Have a good night.
Paul
>The drum is not circular. It is totally different shape.
It is oval or maybe something like a cam on in a car engine ?
At any rate, if we knew "r" as a function of "b" , then maybe we could factor it into the solution.
It is oval or maybe something like a cam on in a car engine ?
At any rate, if we knew "r" as a function of "b" , then maybe we could factor it into the solution.
ASKER
Yes it would be cam like and not much at hand. For now at this phase it will be considered circular. Then as an assignment will be given to some student to improve it to work with a cam like shape.
The string unwrapped produces some rotation. Without the pulley we added to keep string-drum seperation point fixed, we have
String unwrapped = r(angle a + angle b) ' the angle are in radians.
Howevr, if with the pulley in place, there will be no angle b, therefore:
String unwrapped = r(angle a)
The string unwrapped produces some rotation. Without the pulley we added to keep string-drum seperation point fixed, we have
String unwrapped = r(angle a + angle b) ' the angle are in radians.
Howevr, if with the pulley in place, there will be no angle b, therefore:
String unwrapped = r(angle a)
ASKER
The Java program will be given to the student. They have to find mathematical solution for a yet specified cam and handle the required animation.
ASKER
I was very busy with an exam and wasn't able to attend this guestion very regularly. With thw simplified (with small pulley), we can focus on the animation part at https://www.experts-exchange.com/questions/27229753/Animation-plot.html. I will add the latest image with its math function there.
I will close this thread shortly.
Thank you,
Mike
I will close this thread shortly.
Thank you,
Mike
ASKER
Your deserve 5000 point each for educating me. From 500, phoffric gets 400 points for his extensive contribution. Because this part of the question was focusing on the math part not programing portion, 100 points is given to rrz@871311.
Mike
Mike
I see you just closed this while I was putting together some relationships for the original statement.
Since you closed, I just wanted to jot down where I was at. I didn't have time to check it, but here what I have for the relationship between x and angle b:
x = r cos b + (h + r sin b) tan b
Here is a plot on this relationship:
http://www.wolframalpha.com/input/?i=x+%3D+10+cos%28beta%29+%2B+%2860+%2B+10+sin%28beta%29+%29+tan%28beta%29%2C+for+beta+%3D+0+to+.95*pi%2F2
Notice that for larger x, a large change in x results in a small change in angle b, which is what we would expect. As you can see, the max angle b for this problem is pi/2, which occurs when x -> infinity (again, this is what we would expect).
I didn't have time to try to see whether this relationship matches that of rrz@871311. We can certainly try to do that in a related question if desired.
>> String unwrapped = r(angle a + angle b) ' the angle are in radians.
From my other relationships, I do not believe I have this result. We can explore how you got this in a follow-up question if you wish. I have defined, e, the excess rope pull, defined like this: Fix an angle b. Without any wheel rotation, rotate the green rope to be tangent to the wheel - it is green because it's length is still h. Now pull the green rope very slowly (rotating the wheel) until the rope hits the floor (and stop) - the rope is now green. The amount you pulled is length e; and then you have: a = e/r
Since you closed, I just wanted to jot down where I was at. I didn't have time to check it, but here what I have for the relationship between x and angle b:
x = r cos b + (h + r sin b) tan b
Here is a plot on this relationship:
http://www.wolframalpha.com/input/?i=x+%3D+10+cos%28beta%29+%2B+%2860+%2B+10+sin%28beta%29+%29+tan%28beta%29%2C+for+beta+%3D+0+to+.95*pi%2F2
Notice that for larger x, a large change in x results in a small change in angle b, which is what we would expect. As you can see, the max angle b for this problem is pi/2, which occurs when x -> infinity (again, this is what we would expect).
I didn't have time to try to see whether this relationship matches that of rrz@871311. We can certainly try to do that in a related question if desired.
>> String unwrapped = r(angle a + angle b) ' the angle are in radians.
From my other relationships, I do not believe I have this result. We can explore how you got this in a follow-up question if you wish. I have defined, e, the excess rope pull, defined like this: Fix an angle b. Without any wheel rotation, rotate the green rope to be tangent to the wheel - it is green because it's length is still h. Now pull the green rope very slowly (rotating the wheel) until the rope hits the floor (and stop) - the rope is now green. The amount you pulled is length e; and then you have: a = e/r
phoffric, thanks for posting that link. Nice site.
Here is my plot.
http://www.wolframalpha.com/input/?i=sqrt+%28%2810tan%28beta%29+%2B+60%2Fcos%28beta%29%29^2+-+3500%29+-+10+%2C+for+beta+%3D+0+to+.95*pi%2F2
I think your equation has some problems.
When beta = 0 , x = 10 ( shouldn't x = 0)
when beta = 45 degrees = 0.785 rad , x = 74 ( shouldn't that be closer to 60, mine gives 64 )
Here is my plot.
http://www.wolframalpha.com/input/?i=sqrt+%28%2810tan%28beta%29+%2B+60%2Fcos%28beta%29%29^2+-+3500%29+-+10+%2C+for+beta+%3D+0+to+.95*pi%2F2
I think your equation has some problems.
When beta = 0 , x = 10 ( shouldn't x = 0)
when beta = 45 degrees = 0.785 rad , x = 74 ( shouldn't that be closer to 60, mine gives 64 )
My origin was from the center of the circle rather than from the right-most point of the circle. Here is the corrected version:
http://www.wolframalpha.com/input/?i=x+%3D+10+*%281-+cos%28beta%29%29+%2B+%2860+%2B+10+sin%28beta%29+%29+tan%28beta%29%2C+for+beta+%3D+0+to+.95*pi%2F2
http://www.wolframalpha.com/input/?i=x+%3D+10+*%281-+cos%28beta%29%29+%2B+%2860+%2B+10+sin%28beta%29+%29+tan%28beta%29%2C+for+beta+%3D+0+to+.95*pi%2F2
Our two forms differ a little. If we get a follow-on question, we can go over the derivations.
Rope-Drum-Pull.PNG
Rope-Drum-Pull.PNG
My derivation is straight forward.
Using graph above ID:36281780 http:#36281780
Add label P8 to point where dotted green line crosses upper blue horizontal line.
Therefore the length of line P6-P7 = P6-P8 + P8-P7
from trig, we have
(P6-P8) = h/cos(b)
(P8-P7) = r * tan(b)
(P6-P7) = r * tan(b) + h/cos(b)
Using Pythagorean, we have
(P5-P7)^2 = h^2 + (r + x)^2 = (P6-P7)^2 + r^2
finally sub for P6-P7 and solve for x
Using graph above ID:36281780 http:#36281780
Add label P8 to point where dotted green line crosses upper blue horizontal line.
Therefore the length of line P6-P7 = P6-P8 + P8-P7
from trig, we have
(P6-P8) = h/cos(b)
(P8-P7) = r * tan(b)
(P6-P7) = r * tan(b) + h/cos(b)
Using Pythagorean, we have
(P5-P7)^2 = h^2 + (r + x)^2 = (P6-P7)^2 + r^2
finally sub for P6-P7 and solve for x
ASKER
In the attached graph using your e (excess string), if we could establish a relationship between angle b and (h+2) then we can eliminate b from the equation on the graph.
Cam-4.png
Cam-4.png
ASKER
I had some errors and there was some missing info on the last graph. Please consider the following graph.
Also, having y=m(x-x1)+y1 for the string from p2 to p3:
y = -Cot(b)(x-r(1-cos(b)))+(h+
...
Cam-4.png
Also, having y=m(x-x1)+y1 for the string from p2 to p3:
y = -Cot(b)(x-r(1-cos(b)))+(h+
...
Cam-4.png
ASKER
as angle b goes towards zero (say using an adjustable pulley), the lenght of orange r*b also goes to zero. However the purple portion of the string (r*a) remains in play.
Your final equation
> (br + ar + h)^2 = (h + rbSin(b))^2 + (x + r(1 - Cos(b))^2
is wrong. It should be
(br + ar + h)^2 = (h + bSin(b))^2 + (x + r(1 - Cos(b))^2
But I don't see how you could use that equation to work towards a solution.
> (br + ar + h)^2 = (h + rbSin(b))^2 + (x + r(1 - Cos(b))^2
is wrong. It should be
(br + ar + h)^2 = (h + bSin(b))^2 + (x + r(1 - Cos(b))^2
But I don't see how you could use that equation to work towards a solution.
ASKER
I gues you mean:
(br + ar + h)^2 = (h + rSin(b))^2 + (x + r(1 - Cos(b))^2
not
(br + ar + h)^2 = (h + bSin(b))^2 + (x + r(1 - Cos(b))^2
As shown below, we have two equations one with a, b, and x the other one with b, and x. Now, if one of the equations could be solved for b and substituted to the other one, the result will be one equation with a and x variables. Very messy operation but computer may be fine with it.
Cam-4.png
(br + ar + h)^2 = (h + rSin(b))^2 + (x + r(1 - Cos(b))^2
not
(br + ar + h)^2 = (h + bSin(b))^2 + (x + r(1 - Cos(b))^2
As shown below, we have two equations one with a, b, and x the other one with b, and x. Now, if one of the equations could be solved for b and substituted to the other one, the result will be one equation with a and x variables. Very messy operation but computer may be fine with it.
Cam-4.png
ASKER
I forgot to remove b from the term after = sign, should read:
(br + ar + h)^2 = (h + rSin(b))^2 + (x + r(1 - Cos(b))^2 <-- a, b, x
(br + ar + h)^2 = (h + rSin(b))^2 + (x + r(1 - Cos(b))^2 <-- a, b, x
ASKER
Also, the second equation (similarity) should be:
(rCos(b)/(h+rSin(b)))=(rSi
(rCos(b)/(h+rSin(b)))=(rSi
>I gues you mean:
Yes. You are right.
>we might could say yellow and blue triangles are similar
Excellent idea! That leads to a much simipler equation than mine.
>(rCos(b)/(h+rSin(b)))=(rS
Solving for x, we have
x = (h + r*sin(b))tan(b) - r*(1 - cos(b))
The java code show us that it is equivalent to mine.
the results are
b = 000.00 a = 000.00 x = 000.00
b = 000.10 a = 000.03 x = 006.07
b = 000.20 a = 000.12 x = 012.37
b = 000.30 a = 000.29 x = 019.03
b = 000.40 a = 000.54 x = 026.22
b = 000.50 a = 000.88 x = 034.17
b = 000.60 a = 001.35 x = 043.16
b = 000.70 a = 001.99 x = 053.61
b = 000.80 a = 002.84 x = 066.13
b = 000.90 a = 004.01 x = 081.70
b = 001.00 a = 005.66 x = 101.95
b = 001.10 a = 008.09 x = 129.93
b = 001.20 a = 011.93 x = 171.93
b = 001.30 a = 018.73 x = 243.51
b = 001.40 a = 033.70 x = 396.71
b = 001.50 a = 091.42 x = 977.45
The graph (thanks to phoffric for the link) is at
http://www.wolframalpha.com/input/?i=x+%3D+%2860+%2B+10*sin%28beta%29%29*tan%28beta%29+-+10*%281-cos%28beta%29%29%2C+for+beta+%3D+0+to+.95*pi%2F2
"a" function of "b" ,derived from your similar triangle idea, results in the same equation I posted.
Yes. You are right.
>we might could say yellow and blue triangles are similar
Excellent idea! That leads to a much simipler equation than mine.
>(rCos(b)/(h+rSin(b)))=(rS
Solving for x, we have
x = (h + r*sin(b))tan(b) - r*(1 - cos(b))
The java code show us that it is equivalent to mine.
import java.text.DecimalFormat;
public class A {
public static void main(String args[]) {
DecimalFormat df = new DecimalFormat("000.00");
double h = 60;
double r = 10;
double a = 0;
double x = 0;
for(double b = 0; b < 1.58; b += 0.1){
a = (r * Math.tan(b) + h/Math.cos(b) - h)/ r - b;
x = (h + r*Math.sin(b))*Math.tan(b) - r*(1 - Math.cos(b));
System.out.println("b = " + df.format(b) + " a = " + df.format(a) + " x = " + df.format(x));
}
}
}the results are
b = 000.00 a = 000.00 x = 000.00
b = 000.10 a = 000.03 x = 006.07
b = 000.20 a = 000.12 x = 012.37
b = 000.30 a = 000.29 x = 019.03
b = 000.40 a = 000.54 x = 026.22
b = 000.50 a = 000.88 x = 034.17
b = 000.60 a = 001.35 x = 043.16
b = 000.70 a = 001.99 x = 053.61
b = 000.80 a = 002.84 x = 066.13
b = 000.90 a = 004.01 x = 081.70
b = 001.00 a = 005.66 x = 101.95
b = 001.10 a = 008.09 x = 129.93
b = 001.20 a = 011.93 x = 171.93
b = 001.30 a = 018.73 x = 243.51
b = 001.40 a = 033.70 x = 396.71
b = 001.50 a = 091.42 x = 977.45
The graph (thanks to phoffric for the link) is at
http://www.wolframalpha.com/input/?i=x+%3D+%2860+%2B+10*sin%28beta%29%29*tan%28beta%29+-+10*%281-cos%28beta%29%29%2C+for+beta+%3D+0+to+.95*pi%2F2
"a" function of "b" ,derived from your similar triangle idea, results in the same equation I posted.
>> did you notice that angle P6_P5_P1 = complement of angle P6_P7_P4.
I was trying to indicate in http:#36282099 the triangle similarity with this statement. Sorry I wasn't clearer.
You drawing in http:#36300598 very nicely shows many relationships via color-coding. Just a nit in case you are making a presentation.. You probably realize that the orange line (b*r, the one that is tangent to the circle) does not reach the horizontal line.
I was trying to indicate in http:#36282099 the triangle similarity with this statement. Sorry I wasn't clearer.
You drawing in http:#36300598 very nicely shows many relationships via color-coding. Just a nit in case you are making a presentation.. You probably realize that the orange line (b*r, the one that is tangent to the circle) does not reach the horizontal line.
For the animation of your original problem, you can use the results from the Java code. For each increment of time, you can step through the table and look up the values of "a" and "x" for each value of "b".
For the simplified version, the math is much easier.
(h + r*a)^2 = x^2 + h^2
solving for "a" we have
a + sqrt((x/r)^2 + (h/r)^2) - h/r
with h=60 and r=10
a = sqrt((x/10)^2 + 36) - 6
if we take the derivative
http://www.wolframalpha.com/input/?i=derivative+of+sqrt%28%28x%2F10%29^2+%2B+36%29+-+6
Which tells us(if I am thinking correctly) that
(angular velocity of drum) = x/10*sqrt(x^2 + 3600) * (pulling velocity along horizontal)
For the simplified version, the math is much easier.
(h + r*a)^2 = x^2 + h^2
solving for "a" we have
a + sqrt((x/r)^2 + (h/r)^2) - h/r
with h=60 and r=10
a = sqrt((x/10)^2 + 36) - 6
if we take the derivative
http://www.wolframalpha.com/input/?i=derivative+of+sqrt%28%28x%2F10%29^2+%2B+36%29+-+6
Which tells us(if I am thinking correctly) that
(angular velocity of drum) = x/10*sqrt(x^2 + 3600) * (pulling velocity along horizontal)
ASKER
re:> The java code show us that it is equivalent to mine.
My appology if I have missed that.
re:> did you notice that angle P6_P5_P1 = complement of angle P6_P7_P4.
You not about "complement of angle ..." was where I got the idea of triangle similiraty. I guess I wasn't clear indicating that.
re:> You probably realize that the orange line (b*r, the one that is tangent to the circle) does not reach the horizontal line.
You are right. In the attached graph, I have indicated that that portion is not drayn to scale.
I guess, we no longer have to limit ourself to the simplified version with the small pulley. I will post a new question in Java section to see how far we can go.
I am beginner in java, I hope you are willing a code I would be able to execute and wonder how you did it. It will be a very good learning tool for me as well.
I will post the link for the new question here. You are both amazing in math and design ability.
thx
drum-1.png
My appology if I have missed that.
re:> did you notice that angle P6_P5_P1 = complement of angle P6_P7_P4.
You not about "complement of angle ..." was where I got the idea of triangle similiraty. I guess I wasn't clear indicating that.
re:> You probably realize that the orange line (b*r, the one that is tangent to the circle) does not reach the horizontal line.
You are right. In the attached graph, I have indicated that that portion is not drayn to scale.
I guess, we no longer have to limit ourself to the simplified version with the small pulley. I will post a new question in Java section to see how far we can go.
I am beginner in java, I hope you are willing a code I would be able to execute and wonder how you did it. It will be a very good learning tool for me as well.
I will post the link for the new question here. You are both amazing in math and design ability.
thx
drum-1.png
ASKER
here is the link for the new post:
https://www.experts-exchange.com/questions/27237157/Animation-help.html
Thank you,
Mike
https://www.experts-exchange.com/questions/27237157/Animation-help.html
Thank you,
Mike
To work with rate of change in your animation, you will need to determine da/dx even if you cannot get a = a(x) as originally desired. You should provide the Java programmers with the rate of change equations so that they will know how 'a' changes as 'x' changes incrementally.
Now, in previous posts, you do have x = f(b), and you could compute db/dx. To get you started, here is a short video tutorial on this topic:
http://www.khanacademy.org/video/introduction-to-rate-of-change-problems?playlist=Calculus
Likewise, you will be able to compute da/db in a similar (but slightly harder) way.
Good luck in your exam(s)! What are you studying?
Now, in previous posts, you do have x = f(b), and you could compute db/dx. To get you started, here is a short video tutorial on this topic:
http://www.khanacademy.org/video/introduction-to-rate-of-change-problems?playlist=Calculus
Likewise, you will be able to compute da/db in a similar (but slightly harder) way.
Good luck in your exam(s)! What are you studying?
You may be thinking about computing da/dt, and rightly so. But, since, in your animation, dt is a constant (e.g., say 50 ms/frame), then given a rate of, say, dx/dt = 2 inches/second, then your dx is just 2/(1000/50) = .1 inches per frame. Then, if you have dx fixed at .1, and you know da/dx, then you now have da for each position of x.
ASKER
re:> you will need to determine da/dx even if you cannot get a = a(x) as originally desired
For time being, using a timer of some sort, the screen could be refreshed for incremental increase on x. To observer it is as good as animation I think. When a user specifies a rate like 2 inch/sec, the timer parameters could be fine tunn to coorelate to it.
I will handle da/dt or dx/dt when I start using the real configuration of the cam which will replace the circular drum.
re:> short video tutorial
Thank you for this link and also for wolframalpha.com link (very useful).
I am new to java (have been busy wasting time with vb). This question is hobby related and it is not for exam. I just finished my first course in java.
I don't have the necessary skills to do this project at this point. I hope I could get some assitance to putting the basic structure together with EE experts.
Thank you very much. You were very helpful with your contributions.
Mike
For time being, using a timer of some sort, the screen could be refreshed for incremental increase on x. To observer it is as good as animation I think. When a user specifies a rate like 2 inch/sec, the timer parameters could be fine tunn to coorelate to it.
I will handle da/dt or dx/dt when I start using the real configuration of the cam which will replace the circular drum.
re:> short video tutorial
Thank you for this link and also for wolframalpha.com link (very useful).
I am new to java (have been busy wasting time with vb). This question is hobby related and it is not for exam. I just finished my first course in java.
I don't have the necessary skills to do this project at this point. I hope I could get some assitance to putting the basic structure together with EE experts.
Thank you very much. You were very helpful with your contributions.
Mike
ASKER
Paul, the video is very good and easy to follow. It was fairly easy to follow, I enjoyed it.
I am still waiting to see if a Swing expert shows up in your new question over in the Java zone.
Would you consider a solution in a HTML5 canvas ?
http://en.wikipedia.org/wiki/Canvas_element
That might be more fun. I could probably come up with a rough applet that would work first if you want.
First of all, you didn't comment on http:#36304297
Could you please do so?
I was trying to show you( similar to what Khan did in his video), in the simplified case, that taking the derivative would give produce the velocity equation we need for the animation.
For your original problem(on pulley) the more complex equations will preclude such an easy solution. Unless someone suggests a better way, I would use the tabular method that I suggested.
Would you consider a solution in a HTML5 canvas ?
http://en.wikipedia.org/wiki/Canvas_element
That might be more fun. I could probably come up with a rough applet that would work first if you want.
First of all, you didn't comment on http:#36304297
Could you please do so?
I was trying to show you( similar to what Khan did in his video), in the simplified case, that taking the derivative would give produce the velocity equation we need for the animation.
For your original problem(on pulley) the more complex equations will preclude such an easy solution. Unless someone suggests a better way, I would use the tabular method that I suggested.
ASKER
re:> HTML5 canvas ...
Yes it will be just as good. All I want to do is to be able to demo it.
I have some HTML background (and starting HTML5 next october). It will be greatly appreciated.
Please let me know what TA will be most appropriate to post the question?
re:> http:#36304297
I read and was trying to digest and see what direction is better. What is your openion? I am open to have first the simplified version with pulley. And later, on another question we can do the full version without the pulley.
You seem to have your arm around it better than I.
I will wait for TA you want me to add the question so you can doe either of w/ or wo/ pulley using HTML5?
Regards,
Mike
Yes it will be just as good. All I want to do is to be able to demo it.
I have some HTML background (and starting HTML5 next october). It will be greatly appreciated.
Please let me know what TA will be most appropriate to post the question?
re:> http:#36304297
I read and was trying to digest and see what direction is better. What is your openion? I am open to have first the simplified version with pulley. And later, on another question we can do the full version without the pulley.
You seem to have your arm around it better than I.
I will wait for TA you want me to add the question so you can doe either of w/ or wo/ pulley using HTML5?
Regards,
Mike
If you hit the "ask a related question" link when creating a new question, then the Q's are tied together - makes for a better PAQ.
This Question is taking forever to load (on an older PC, a P4, and only 1GB Ram; 2.8 GHz; single core, but hyperthreaded). Must be all those nice drawings.
This Question is taking forever to load (on an older PC, a P4, and only 1GB Ram; 2.8 GHz; single core, but hyperthreaded). Must be all those nice drawings.
I have been fooling around with Java for 10 years. I have made some applets and JFrame apps. I am a JSP expert. But I am not an expert in Java Swing or HTML5. So it will be a lot of work for me either way.
To try get help in HTML5 canvas, post in
https://www.experts-exchange.com/Programming/Languages/Scripting/JavaScript/
But before you ask for a coding expert, you should figure out the math. Do you understand what I posted in http:#36304297 ? If not we can discuss it here.
To try get help in HTML5 canvas, post in
https://www.experts-exchange.com/Programming/Languages/Scripting/JavaScript/
But before you ask for a coding expert, you should figure out the math. Do you understand what I posted in http:#36304297 ? If not we can discuss it here.
By the way, you still have an error on your graph at
https://www.experts-exchange.com/questions/27237157/Animation-help.html
You have rbSin(b) but it should labeled rSin(b) .
https://www.experts-exchange.com/questions/27237157/Animation-help.html
You have rbSin(b) but it should labeled rSin(b) .
ASKER
re:> You have rbSin(b) but it should labeled rSin(b) .
You are right, I just corrected it. I will use the revised graph next time I post this question.
re:> To try get help in HTML5 canvas, post in
I will add a new question in both Javascript and Java sections with a better description crlaifying math portion the best way I can.
re:> post http:#36304297
There are following items discussed in it:
- Would you consider a solution in a HTML5 canvas ?: Yes HTML5 will be fine. Using wikipedia link you have included, I tested it via the sample codes included there. My explorer doesn't work with HTML5 but my Google Chrome browser seems to handle HTML5. With Chrome the following code, nevertheless, didn't draw the expected rectangle.
Wiki says--> Using JavaScript, you can draw on the canvas...
var example = document.getElementById('e
var context = example.getContext('2d');
context.fillStyle = "rgb(255,0,0)";
context.fillRect(30, 30, 50, 50);
but it didn't draw. I suspect, its execution is not like pasting it on a note-pad and saving it as Test.html, for example, and openning it with Google Chrome. This works for HTML4 apparently not for HTML5
-------------------
- as far as having it with the pulley or with out it, either way is fine with me (although I prefer without pulley). Whatever that would be done easier and in a lesser time.
re:> http:#36304297 -For the animation of your original problem, you can use the results from the Java code. For each increment of time...
This sounds is a good short cut where we draw in time increments, advancing say angle b to get x and then use hese info to get angle a. Or better yet, complie a table as you did once before at ID:36283363.
re:> of course if we use simplified version then we just increment x and get angle a directly.
Eventually I will do away with simplyfied version. But it is okay for now to start with simplyfied version.
- with or without pulley, do you want me to take a derivitive? Or you will do it yourself. Over all I understand what you are talking about but I havn't gone through it and derive it on my own. It is your choise, whatever progress we make I am happy with it.
re:> "ask a related question"
Thank you for the info. This is a bit new feature and it is good to start using it.
re:> Must be all those nice drawings.... slow response:
We are about done with this thread. Thank you for your passion (I mean both experts).
The new link is: https://www.experts-exchange.com/questions/27237703/Animation-help-2.html
My TA listing was not showing clearly. I nee to sor them out.
You are right, I just corrected it. I will use the revised graph next time I post this question.
re:> To try get help in HTML5 canvas, post in
I will add a new question in both Javascript and Java sections with a better description crlaifying math portion the best way I can.
re:> post http:#36304297
There are following items discussed in it:
- Would you consider a solution in a HTML5 canvas ?: Yes HTML5 will be fine. Using wikipedia link you have included, I tested it via the sample codes included there. My explorer doesn't work with HTML5 but my Google Chrome browser seems to handle HTML5. With Chrome the following code, nevertheless, didn't draw the expected rectangle.
Wiki says--> Using JavaScript, you can draw on the canvas...
var example = document.getElementById('e
var context = example.getContext('2d');
context.fillStyle = "rgb(255,0,0)";
context.fillRect(30, 30, 50, 50);
but it didn't draw. I suspect, its execution is not like pasting it on a note-pad and saving it as Test.html, for example, and openning it with Google Chrome. This works for HTML4 apparently not for HTML5
-------------------
- as far as having it with the pulley or with out it, either way is fine with me (although I prefer without pulley). Whatever that would be done easier and in a lesser time.
re:> http:#36304297 -For the animation of your original problem, you can use the results from the Java code. For each increment of time...
This sounds is a good short cut where we draw in time increments, advancing say angle b to get x and then use hese info to get angle a. Or better yet, complie a table as you did once before at ID:36283363.
re:> of course if we use simplified version then we just increment x and get angle a directly.
Eventually I will do away with simplyfied version. But it is okay for now to start with simplyfied version.
- with or without pulley, do you want me to take a derivitive? Or you will do it yourself. Over all I understand what you are talking about but I havn't gone through it and derive it on my own. It is your choise, whatever progress we make I am happy with it.
re:> "ask a related question"
Thank you for the info. This is a bit new feature and it is good to start using it.
re:> Must be all those nice drawings.... slow response:
We are about done with this thread. Thank you for your passion (I mean both experts).
The new link is: https://www.experts-exchange.com/questions/27237703/Animation-help-2.html
My TA listing was not showing clearly. I nee to sor them out.
>This sounds is a good short cut
It's not a short cut. It's a work around.
I'll start working on the programming in the morning. I don't know how long it will take me to come up with something. I'll start with the HTML5 canvas.
It's not a short cut. It's a work around.
I'll start working on the programming in the morning. I don't know how long it will take me to come up with something. I'll start with the HTML5 canvas.
ASKER
rrz@871311,
There is no dealine. I appreciate for your follow up on this.
Regards,
Mike
There is no dealine. I appreciate for your follow up on this.
Regards,
Mike
ASKER
rrz@871311,
I now realize the importance of of your code at ID:36304001. Sorry for being so slow about it.
Mike
I now realize the importance of of your code at ID:36304001. Sorry for being so slow about it.
Mike
ASKER
L = the length of the arch from p1 to p2 = b*r
m = r-(r*cos(b))
k = r*sin(b)
(h+L)^2 = (h + k)^2 + (x + m)^2
Please see the attached image below for angle b and lenghts m and k.
Mike
Cam-2a.png