Mike Eghtebas

asked on

I have following two equations (r and h are constants):

x=x(b)=r (cos(b)+ tan(b)sin(b)-1)+ tan(b)h (eq. 1)

a =a(x,b)= (1/r)*sqrt((h + rsin(b))^2 + (x + r - rcos(b))^2) - b-h/r (eg. 2)

to make it readable, if we assume S=Sin(b), C=Cos(b), and T=tan(b), then we get:

x=x(b)=r (C+ TS -1)+ Th <-- these forms may be useful may be not

a =a(x,b)= (1/r)*sqrt((h + rS)^2 + (x + r - rC)^2) - b-h/r;

I want to have the following functions if possible:

a(x)=....? (eq. 3)

b(x)=....? (eq. 4) <-- this one is not that important.

Question: Is it possible to find a(x) and b(x) having (eq. 1) and (eq. 4) above?

for background discussions please see: https://www.experts-exchange.com/questions/27237703/Animation-help-2.html

Thank you

x=x(b)=r (cos(b)+ tan(b)sin(b)-1)+ tan(b)h (eq. 1)

a =a(x,b)= (1/r)*sqrt((h + rsin(b))^2 + (x + r - rcos(b))^2) - b-h/r (eg. 2)

to make it readable, if we assume S=Sin(b), C=Cos(b), and T=tan(b), then we get:

x=x(b)=r (C+ TS -1)+ Th <-- these forms may be useful may be not

a =a(x,b)= (1/r)*sqrt((h + rS)^2 + (x + r - rC)^2) - b-h/r;

I want to have the following functions if possible:

a(x)=....? (eq. 3)

b(x)=....? (eq. 4) <-- this one is not that important.

Question: Is it possible to find a(x) and b(x) having (eq. 1) and (eq. 4) above?

for background discussions please see: https://www.experts-exchange.com/questions/27237703/Animation-help-2.html

Thank you

Last Comment

ASKER

re:> What is b(x)? I don't see b shown as a function anywhere

lets ignore b(x). But, if we had it in a much simpler form say x=rcos(b) then

b=ArcCos(x/r) and now we have b(x)=ArcCos(x/r)

re:> Also, if a is defined as taking two parameters then how do you drop one? How do you get from a(x,b) ...

This is why I am posting the question here hoping someone familiar with chain function possibly could come up with the answer.

thx

lets ignore b(x). But, if we had it in a much simpler form say x=rcos(b) then

b=ArcCos(x/r) and now we have b(x)=ArcCos(x/r)

re:> Also, if a is defined as taking two parameters then how do you drop one? How do you get from a(x,b) ...

This is why I am posting the question here hoping someone familiar with chain function possibly could come up with the answer.

thx

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Okay I get it. So b(x) actually is important because you replace b in the a(x,b) equation with b(x). I've never seen equations expressed as functions, but it makes sense.

I verified pkwan's math and it looks good. He used the trig identities S^2 + C^2 = 1 and S/C = T to derive eq 6 and the quadratic formula to derive eqs 7 and 8. The rest is just pretty simple algebra. Very well done.

Don't forget that when you introduce an ArcCos, you need to constrain the domain.

I verified pkwan's math and it looks good. He used the trig identities S^2 + C^2 = 1 and S/C = T to derive eq 6 and the quadratic formula to derive eqs 7 and 8. The rest is just pretty simple algebra. Very well done.

Don't forget that when you introduce an ArcCos, you need to constrain the domain.

When x = 0, a should be 0...

====

>> x = 10 (cos(b)+ tan(b)sin(b)-1)+ tan(b)60

When b = 0, then x is 0 and vice-versa

http://www.wolframalpha.com/input/?i=10+%28cos%28b%29%2B+tan%28b%29sin%28b%29-1%29%2B+tan%28b%2960+for+b%3D0+to+pi%2F2

====

>> a(0,b) = (1/10)*sqrt((60 + 10sin(b))^2 + (0 + 10 - 10cos(b))^2) - b-60/10

When b and x are 0, a is 0

http://www.wolframalpha.com/input/?i=%281%2F10%29*sqrt%28%2860+%2B+10sin%280%29%29%5E2+%2B+%280+%2B+10+-+10cos%280%29%29%5E2%29+-+0-60%2F10

====

But, ...

>> a = (1/r)*sqrt((x+r)^2+h^2-r^2) -h/r - ArcCos([r(x+r) - h sqrt(-r^2+(x+r)^2+h^2)] / [(x+r)^2+h^2]) ... (eq. 11)

So, for equivalency, if x = 0, then a should be 0, but..

http://www.wolframalpha.com/input/?i=%281%2F10%29*sqrt%28%28x%2B10%29%5E2%2B60%5E2-10%5E2%29+-60%2F10+-+ArcCos%28%5B10%28x%2B10%29+-++60+sqrt%28-10%5E2%2B%28x%2B10%29%5E2%2B60%5E2%29%5D+%2F+%5B%28x%2B10%29%5E2%2B60%5E2%5D%29%2C+for+x+%3D+0+to+45

====

On another note, not related to this question, when looking at the graphs, it appears that the relationship between a and b is suspect, since for large x, then a small change in b should result in a large change in a. See related question: http://rdsrc.us/qNFw9P

====

>> x = 10 (cos(b)+ tan(b)sin(b)-1)+ tan(b)60

When b = 0, then x is 0 and vice-versa

http://www.wolframalpha.com/input/?i=10+%28cos%28b%29%2B+tan%28b%29sin%28b%29-1%29%2B+tan%28b%2960+for+b%3D0+to+pi%2F2

====

>> a(0,b) = (1/10)*sqrt((60 + 10sin(b))^2 + (0 + 10 - 10cos(b))^2) - b-60/10

When b and x are 0, a is 0

http://www.wolframalpha.com/input/?i=%281%2F10%29*sqrt%28%2860+%2B+10sin%280%29%29%5E2+%2B+%280+%2B+10+-+10cos%280%29%29%5E2%29+-+0-60%2F10

====

But, ...

>> a = (1/r)*sqrt((x+r)^2+h^2-r^2

So, for equivalency, if x = 0, then a should be 0, but..

http://www.wolframalpha.com/input/?i=%281%2F10%29*sqrt%28%28x%2B10%29%5E2%2B60%5E2-10%5E2%29+-60%2F10+-+ArcCos%28%5B10%28x%2B10%29+-++60+sqrt%28-10%5E2%2B%28x%2B10%29%5E2%2B60%5E2%29%5D+%2F+%5B%28x%2B10%29%5E2%2B60%5E2%5D%29%2C+for+x+%3D+0+to+45

====

On another note, not related to this question, when looking at the graphs, it appears that the relationship between a and b is suspect, since for large x, then a small change in b should result in a large change in a. See related question: http://rdsrc.us/qNFw9P

ASKER

Hi pkwan,

Thank you for the solution. I included the functions in a java program shown below to test the equations.

At x=0, I was expecting to get zero for both a and b, but I am getting:

angle a angel b x

-02.81 02.81 00.00

-02.79 02.80 00.25

-02.78 02.80 00.50

-00.01 02.26 23.75

00.03 02.25 24.00

00.07 02.25 24.25

00.11 02.25 24.50

00.15 02.24 24.75

00.19 02.24 25.00

06.26 01.94 59.50

06.30 01.93 59.75

Most likely this dicrepency is due to my mistkes converting from your math equation to java equations shown below:

I will go over it to locate my mistakes.

Thank you for the solution. I included the functions in a java program shown below to test the equations.

At x=0, I was expecting to get zero for both a and b, but I am getting:

angle a angel b x

-02.81 02.81 00.00

-02.79 02.80 00.25

-02.78 02.80 00.50

-00.01 02.26 23.75

00.03 02.25 24.00

00.07 02.25 24.25

00.11 02.25 24.50

00.15 02.24 24.75

00.19 02.24 25.00

06.26 01.94 59.50

06.30 01.93 59.75

Most likely this dicrepency is due to my mistkes converting from your math equation to java equations shown below:

I will go over it to locate my mistakes.

```
import java.text.DecimalFormat;
public class Data1 {
public static void main(String args[]) {
DecimalFormat df = new DecimalFormat("00.00");
double h = 30;
double r = 5;
double a = 0;
double b = 0;
System.out.println("angle a\t angel b x");
for(double x = 0; x < 60; x +=0.25){
//b = ArcCos([r(x+r) + h sqrt(-r^2+(x+r)^2+h^2)] / [(x+r)^2+h^2]) ................... (eq. 7)
b = Math.acos((r*(x+r) - h*Math.sqrt(-Math.pow(r,2)+Math.pow((x+r),2)+
Math.pow(h,2))) / (Math.pow((x+r),2)+Math.pow(h,2)));
// a = (1/r)*sqrt((x+r)^2+h^2-r^2) -h/r - ArcCos([r(x+r) - h sqrt(-r^2+(x+r)^2+h^2)] / [(x+r)^2+h^2]) ... (eq. 11)
a=(1/r)*Math.sqrt(Math.pow((x+r),2)+Math.pow(h,2)-Math.pow(r,2))-(h/r)-
Math.acos((r*(x+r)-h*Math.sqrt(-1*Math.pow(r,2)+Math.pow((x+r),2)+
Math.pow(h,2)))/((Math.pow((x+r),2)+Math.pow(h,2))));
// System.out.println(" " + df.format(a*(180/Math.PI)) + " " + df.format(b*(180/Math.PI)) + " " + df.format(x));
System.out.println(" " + df.format(a) + " " + df.format(b) + " " + df.format(x));
}
}
}
```

ASKER

Hi phoffric,

I didn't see your post.

I didn't see your post.

ASKER

I should have used:

x=x(b)=r (cos(b)+ tan(b)sin(b)-1)+ tan(b)h (eq. 1)

a = (r *tan(b) + h/cos(b) - h)/r - b (eg. 2) much simpler

x = r (C+T S -1)+T/h (eq. 1a)

a = (r T + h/C - h)/r - b (eg. 2a)

now will try the solution from pkwan on these two equations,

brb

x=x(b)=r (cos(b)+ tan(b)sin(b)-1)+ tan(b)h (eq. 1)

a = (r *tan(b) + h/cos(b) - h)/r - b (eg. 2) much simpler

x = r (C+T S -1)+T/h (eq. 1a)

a = (r T + h/C - h)/r - b (eg. 2a)

now will try the solution from pkwan on these two equations,

brb

ASKER

pkwan,

I wonder if you have some time to take a look at this:

https://www.experts-exchange.com/questions/27242530/have-x-b-to-find-b-x.html

Thank you,

Mike

I wonder if you have some time to take a look at this:

https://www.experts-exchange.com/questions/27242530/have-x-b-to-find-b-x.html

Thank you,

Mike

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Also, if a is defined as taking two parameters then how do you drop one? How do you get from a(x,b) to just a(x)? The only way I could think of where that would make sense would be if b cancelled out of the equation, but it's not going to.

We need a lot more info before we can answer this question.