Crank/Piston Motion: Understanding dB/dt, dB/dθ & d2B/dT2

[tomc612]

Hi,
Looking for some help on a crank/piston motion equation..

If a crank rotates counter clockwise measured in Radians/sec -as d\theta/dt = K T = time in secs

The arms of the crank A and B are fixed and B is the stroke length with \theta as the angle between A and B
So far I've got that B = Acos\theta + \sqrt{C^2 -(Asin\theta)^2}

Min/max for B is at \theta = 0, Pi, 2Pi.

Q. What is the relationship between dB/dt and dB/d\theta -
A. is this that dB/dT = d\theta/dT x dB/d\theta?

Q. obtain a formula for dB/dT. Answer should be in terms of A, C, \theta and K
A. dB/d\theta = -Asin\theta+ A^2sin\thetacos\theta/\sqrt{C^2 -(Asin\theta)^2}

So my question is dB/dt then simply the the above formula for dB/d\theta x K
=K(-Asin\theta+ A^2sin\thetacos\theta/\sqrt{C^2 -(Asin\theta)^2})

Q. Obtain a formula for d^2B/dT^2
A. if K is constant is that then second derivative of B x K
= B".K
Haven't got to working out the second derivative of B just yet,

any advice appreciated
Thanks
TomView attachment 6151

 

[MarkFL]

Hello Tom,

I think what I would do here is first use the Law of Cosines to determine $b$:

\(\displaystyle c^2=a^2+b^2-2ab\cos(\theta)\)

Arrange in standard quadratic form:

\(\displaystyle b^2-2a\cos(\theta)b+a^2-c^2=0\)

Apply the quadratic formula, and discard the negative root (since we must have $c>a$):

\(\displaystyle b=\frac{2a\cos(\theta)+\sqrt{4a^2\cos^2(\theta)+4(c^2-a^2)}}{2}=a\cos(\theta)+\sqrt{c^2-a^2\sin^2(\theta)}\)

This agrees with your result.

Q. What is the relationship between \(\displaystyle \d{b}{t}\) and \(\displaystyle \d{b}{\theta}\)?

Using the chain rule, we may state:

\(\displaystyle \d{b}{t}=\d{b}{\theta}\cdot\d{\theta}{t}\)

We are given:

\(\displaystyle \d{\theta}{t}=k\)

Therefore, we obtain:

\(\displaystyle \d{b}{t}=k\d{b}{\theta}\)

This is where you were headed, but you didn't make the substitution for $k$.

Q. Obtain a formula for \(\displaystyle \d{b}{t}\). Answer should be in terms of $a$, $c$, $\theta$ and $k$.

Okay, given that we found:

\(\displaystyle b(\theta)=a\cos(\theta)+\sqrt{c^2-a^2\sin^2(\theta)}\)

Differentiating w.r.t $\theta$, there results:

\(\displaystyle \d{b}{\theta}=-a\sin(\theta)+\frac{1}{2}(c^2-a^2\sin^2(\theta))^{-\frac{1}{2}}(-2a^2\sin(\theta)\cos(\theta))=-a\sin(\theta)\left(1+\frac{a\cos(\theta)}{\sqrt{c^2-a^2\sin^2(\theta)}}\right)\)

And so, using our result from the first question, we have:

\(\displaystyle \d{b}{t}=-ak\sin(\theta)\left(1+\frac{a\cos(\theta)}{\sqrt{c^2-a^2\sin^2(\theta)}}\right)\)

Now, before we move on to the final question, we can observe that our computation will be simpler if we write:

\(\displaystyle \d{b}{t}=\frac{abk\sin(\theta)}{a\cos(\theta)-b}\)

Now we may apply the quotient and chain rules as appropriate. :D

 

[tomc612]

Hi Thanks for the help, glad to see I wasn't too far off.

One thing for the formula for B, i reached the formula by creating 'h' to represent the height of the joint of A+C, and then split B in B1 and B2 at h to create to right angle triangles.

B=B1 + B2
B1 = Acos\theta
B2 = \sqrt{C^2 -h^2}
h = Asin\theta
B2 =\sqrt{C^2 -(Asin\theta)^2}
b = Acos\theta +\sqrt{C^2 -(Asin\theta)^2}

 

[tomc612]

Hello Tom,

I think what I would do here is first use the Law of Cosines to determine $b$:

\(\displaystyle c^2=a^2+b^2-2ab\cos(\theta)\)

Arrange in standard quadratic form:

\(\displaystyle b^2-2a\cos(\theta)b+a^2-c^2=0\)

Apply the quadratic formula, and discard the negative root (since we must have $c>a$):

\(\displaystyle b=\frac{2a\cos(\theta)+\sqrt{4a^2\cos^2(\theta)+4(c^2-a^2)}}{2}=a\cos(\theta)+\sqrt{c^2-a^2\sin^2(\theta)}\)

This agrees with your result.

Q. What is the relationship between \(\displaystyle \d{b}{t}\) and \(\displaystyle \d{b}{\theta}\)?

Using the chain rule, we may state:

\(\displaystyle \d{b}{t}=\d{b}{\theta}\cdot\d{\theta}{t}\)

We are given:

\(\displaystyle \d{\theta}{t}=k\)

Therefore, we obtain:

\(\displaystyle \d{b}{t}=k\d{b}{\theta}\)

This is where you were headed, but you didn't make the substitution for $k$.

Q. Obtain a formula for \(\displaystyle \d{b}{t}\). Answer should be in terms of $a$, $c$, $\theta$ and $k$.

Okay, given that we found:

\(\displaystyle b(\theta)=a\cos(\theta)+\sqrt{c^2-a^2\sin^2(\theta)}\)

Differentiating w.r.t $\theta$, there results:

\(\displaystyle \d{b}{\theta}=-a\sin(\theta)+\frac{1}{2}(c^2-a^2\sin^2(\theta))^{-\frac{1}{2}}(-2a^2\sin(\theta)\cos(\theta))=-a\sin(\theta)\left(1+\frac{a\cos(\theta)}{\sqrt{c^2-a^2\sin^2(\theta)}}\right)\)

And so, using our result from the first question, we have:

\(\displaystyle \d{b}{t}=-ak\sin(\theta)\left(1+\frac{a\cos(\theta)}{\sqrt{c^2-a^2\sin^2(\theta)}}\right)\)

Now, before we move on to the final question, we can observe that our computation will be simpler if we write:

\(\displaystyle \d{b}{t}=\frac{abk\sin(\theta)}{a\cos(\theta)-b}\)

Now we may apply the quotient and chain rules as appropriate. :D

Hi, Also need some help on a following question to this same problem.

need work on optimisation for the values of A+B

I tried optimising by implicit differentiation of a right and angled triangle, however the values I get are out side of the given range

12. we change the values of a and c, subject to the following requirements:
a + c = 30; and 2\le  c- 􀀀 \lea  10:
(a) What is the smallest possible value for a? What is the largest possible value for a?
Justify.

 

[MarkFL]

What is this line supposed to say?

...a + c = 30; and 2\le  c- 􀀀 \lea  10:...

 

[tomc612]

What is this line supposed to say?

Sorry, the question should read:

View attachment 6187

 

[MarkFL]

Okay, we are given:

\(\displaystyle c=30-a\)

And:

\(\displaystyle 2\le c-a\le10\)

Substitute for $c$:

\(\displaystyle 2\le 30-a-a\le10\)

Simplify:

\(\displaystyle 2\le 30-2a\le10\)

You should be able to algebraically show this is equivalent to:

\(\displaystyle 10\le a\le14\)