MHB 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 $b$:

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

Arrange in standard quadratic form:

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

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

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

This agrees with your result.

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

Using the chain rule, we may state:

$$\d{b}{t}=\d{b}{\theta}\cdot\d{\theta}{t}$$

We are given:

$$\d{\theta}{t}=k$$

Therefore, we obtain:

$$\d{b}{t}=k\d{b}{\theta}$$

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

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

Okay, given that we found:

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

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

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

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

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

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

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

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 $b$:

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

Arrange in standard quadratic form:

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

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

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

This agrees with your result.

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

Using the chain rule, we may state:

$$\d{b}{t}=\d{b}{\theta}\cdot\d{\theta}{t}$$

We are given:

$$\d{\theta}{t}=k$$

Therefore, we obtain:

$$\d{b}{t}=k\d{b}{\theta}$$

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

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

Okay, given that we found:

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

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

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

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

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

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

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

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:

$$c=30-a$$

And:

$$2\le c-a\le10$$

Substitute for $c$:

$$2\le 30-a-a\le10$$

Simplify:

$$2\le 30-2a\le10$$

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

$$10\le a\le14$$