Calculate crank angle given pushrod displacement

**russiver** · Jul 22nd 2014, 06:58 AM

This looks like it should be fairly straightforward on paper, but ...

Looking to calculate the angle of crank rotation (alpa), about pivot point, given linear displacement of pushrod of 30 units.

Knowns:

Pushrod length = 150
Crank length = 40
Pushrod displacement = 30

Thus:

**ebaines** · Jul 22nd 2014, 09:08 AM

You can proceed as follows, using the dimensions I added to your figure:

1. The x, y coordinates of the tip of the crank are (R sin alpha, R cos alpha)
2. Dimension h is R-y
3. Dimension d is sqrt(R^2-h^2)
4. Finally, dimension D is L+x-d, or:

$D = L - R \sin \alpha - \sqrt {L^2-R^2 (1-\cos \alpha)^2}$

Now the problem is how to make the angle alpha the independent variable - it's a mess, but after some manipulation I get the folowing quadratic equation:

$(c^2+1)sin^2 \alpha + 2kc \sin \alpha + (k^2+1) = 0$

where c and k are:

$c = \frac {L-D} R$

$k = \frac {D^2/2-LD+R^2}{R^2}$

This will yield two values for alpha. Take the arcsin to find the corresponding angles, but as always with arcsins be careful to understand which quadrant alpha belongs in.

Calculate crank angle given pushrod displacement-crank2.jpg

**HallsofIvy** · Jul 22nd 2014, 10:51 AM

Another way if looking at it: If we set up and xy-coordinate system with origin at the center of the wheel, x axis horizontal and y axis vertical, when the crank arm makes angle $\theta$ with the positive x axis, the end of the crank arm is at $(40 cos(\theta), 40 sin(\theta))$ . The push rod is a straight line of length 150 from [tex](40 cos(\theta), 40 sin(\theta))[tex] to the point (-150+ 30, 0)= (-120, 0). So we have length squared equal to $(-120- 40cos(\theta))^2+ (40 sin(\theta))^2= 150^2$ . That reduces to $9600 cos(\theta)= 2500$ .

**ebaines** · Jul 22nd 2014, 11:51 AM

Originally Posted by HallsofIvy

$(-120- 40cos(\theta))^2+ (40 sin(\theta))^2= 150^2$ .

I think this should be: $(-120- 40cos(\theta))^2+ (40-40 sin(\theta))^2= 150^2$

The answer I get for D=30 is theta = 47.4 degress, or alpha = 90-theta = 42.6 degrees.

**russiver** · Jul 22nd 2014, 12:53 PM

Thanks for looking into this. I'm still a little lost as neither approach above gets me in the ball park.

47.4 degrees looks about right for alpha, not theta (drawing this out to scale)?

Can you please just clarify what R is in your equations above? I assume R=40, but then I'm a little lost with equation (3) Dimension d is sqrt(R^2-h^2)? Is R here not 150?

**ebaines** · Jul 22nd 2014, 01:23 PM

Originally Posted by russiver

47.4 degrees looks about right for alpha, not theta

Right - sorry I got that backwards. Shoukld be: alpha = 47.4 degress, or theta= 90-theta = 42.6 degrees.

Originally Posted by russiver

Can you please just clarify what R is in your equations above? I assume R=40...

Yes, R=radius of the circle the crank turns in, or 40. I should have made that clearer. Also L = length of the pushrod =150.

Originally Posted by russiver

Dimension d is sqrt(R^2-h^2)? Is R here not 150?

Sorry, that's another typo. Thanks for catching these errors - I must be getting tired! Equation 3 in my original response should be: Dimension d is sqrt(L^2-h^2).

By the way, there's another solution: Alpha=169.5 degrees also results in a displacement of 30.

**russiver** · Jul 22nd 2014, 01:57 PM

With you now. Thanks again for working this through.

Just need to figure out how you got to $(c^2+1)sin^2 \alpha + 2kc \sin \alpha + (k^2+1) = 0$ from $D = L - R \sin \alpha - \sqrt {L^2-R^2 (1-\cos \alpha)^2}$ .

My algebra is pretty rusty, but I'm hoping I can get there. Getting rid of $cos \alpha$ is my biggest concern. I'm thinking (probably wrong) a trig identity is needed?

**HallsofIvy** · Jul 22nd 2014, 01:59 PM

Originally Posted by ebaines

I think this should be: $(-120- 40cos(\theta))^2+ (40-40 sin(\theta))^2= 150^2$

Yes, I forgot that the center of the wheel was 40 cm below the tip of the pushrod. Thanks.

The answer I get for D=30 is theta = 47.4 degress, or alpha = 90-theta = 42.6 degrees.

**ebaines** · Jul 23rd 2014, 04:54 AM

The algebra is messy. First get the square root term by itself, then square both sides. You'll end up with terms of sin^2, sin, and cos and cos^2. You can then replace cos^2 alpha + sin^2 alpaha with 1, and replace the remaining cosine term with sqrt(1-sin^2). Rearrange to get that square root by itself, square both sides, do some final rearrangement, and you're done!

**russiver** · Jul 23rd 2014, 05:00 AM

Thanks. I''l give this a try and see how I get on.

**russiver** · Jul 23rd 2014, 06:58 AM

Ok so we start with:

$D = L - R sin \alpha - \sqrt {L^2-R^2 (1- cos \alpha)^2}$

Isolating square root term:

$\sqrt {L^2-R^2 (1- cos \alpha)^2}= L - Rsin \alpha -D$

Squaring both sides:

${L^2-R^2 (1- cos \alpha)^2} = L^2 - R^2sin^2\alpha - D^2$

Dividing by R²:

$\frac {L^2} {R^2}-(1- cos \alpha)^2 = \frac {L^2-D^2} {R^2} - sin^2\alpha$

Expanding brackets:

$\frac {L^2} {R^2}- 1 + 2cos\alpha + cos^2 \alpha = \frac {L^2-D^2} {R^2} - sin^2\alpha$

Replacing $cos^2 \alpha + sin^2 \alpha$ with 1:

$\frac {L^2} {R^2}- 1 + 2cos\alpha + 1 = \frac {L^2-D^2} {R^2}$

Am I good so far? Concerned I'm going to lose L?

**russiver** · Jul 23rd 2014, 09:50 AM

Can already see that I messed up removing the square root - squaring both sides. Apologies algebra and I parted company some years ago!

**ebaines** · Jul 23rd 2014, 10:57 AM

First - my apologies, but there was another typo on my part. The correct formula is:

$D= L+R \sin \alpha - \sqrt{L^2-R^2(1-\cos \alpha)^2}$

Rearrange to get:

$L+R \sin \alpha-D= \sqrt{L^2-R^2(1-\cos \alpha)^2}$

and then square both sides. The square of $L+R \sin \alpha -D$ is:

$(L+R \sin \alpha -D)((L+R \sin \alpha -D) = L^2 + R^2 \sin^2 \alpha + D^2 +2LR \sin \alpha -2LD -2DR \siin \alpha$

So we have:

$L^2 + R^2 \sin^2 \alpha + D^2 +2LR \sin \alpha -2LD -2DR \sin \alpha = L^2-R^2(1-2 \cos \alpha + \cos^2 \alpha)$

The L^2 terms cancel, and if we move the R^2cos^2 term to the left and combine the sine terms we get:

$R^2 (\sin^2 \alpha + \cos^2 \alpha) +D^2 -2LD + 2R(L-D) \sin \alpha = -R^2 +2R^2 \cos \alpha$

$= 2R^2 (1- \cos \alpha) +D^2-2LD +2R(L-D) \sin \alpha = 0$

Divide through by 2R^2, and rearrange:

$1+\frac {D^2}{2R^2}-\frac {LD}{R^2} = \cos \alpha - (\frac {L-D} R) \sin \alpha$

Here is where we set $c = \frac {L-D} R$ and $k = 1+\frac {D^2}{2R^2}-\frac {LD}{R^2}$ , and replace the cosine with $\cos \alpha = \sqrt{1- \sin^2 \alpha}$ :

$\sqrt{1- \sin^2 \alpha} = k + c \sin \alpha$

Square both sides:

$1 - \sin^2 \alpha = k^2 + 2kc \sin \alpha + c^2 \sin^2 \alpha$

$0 = (c^2+1) \sin^2 \alpha + 2kc \sin \alpha + (k^2-1)$

This is a quadratic in sin alpha. Use the quadratic formula to find two values of sin alpha. Take the arc sine to find the angles. When you do this you should get:

$\alpha_1 = 47.4^{\circ}, \ \alpha_2 = 10.5^{\circ}$

As I noted previously we need to be careful about making sure alpha s in the correct quadrant. That second value for alpha should actually be 180-10.5 = 169.5.

**russiver** · Jul 23rd 2014, 02:06 PM

Big thanks again for working this through and walking me through the route to it.

Originally Posted by ebaines

As I noted previously we need to be careful about making sure alpha s in the correct quadrant. That second value for alpha should actually be 180-10.5 = 169.5.

Just one final question, without visualising (sketching out) the solutions, is the anyway to deduce the correct quadrants for both solutions? I know sine is positive in the first and second quadrant, but just wonder how you figured the second solution 10.5 and not the first 47.4 needed subtracting from 180 without sketching?

**ebaines** · Jul 24th 2014, 04:29 AM

When evaluating an arcsine there is no "automatic" way to know which is the correct quadrant - you have to consider the geometry and see which solution fits. In this case you could calculate the length of displacement D for each of the possible angles: 47.4, 10.5,180-47.4, and 180-10.5 to see which one(s) result in D= 30.

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