Mechanical Advantage in a crank system

In summary: Yes, that is probably right,as you can see the Lever is actually in a v type shape, my estimations of angle were perhaps not the best.I can calculate the torque on the crank with a perpendicular force of 186 N applied as pictured at θ 55 degrees,to T= 21.34 Nm.And at this moment the distance X perpendicular from the crank to the bottom cylinder = 70.7mmbut what is the perpendicular force applied on the cylinder by H at this moment?How does one calculate mechanical advantage?The ApplicationThe application is a lever espresso machine.
  • #1
powdernose
12
0
I am not a student btw,
this is seeking extra knowledge regarding a hobby of mine, and it has been too long since I've tried to solve a physics problem.

1. Homework Statement
caravel.png


I have the above system in which Lever L (200mm) applies force to a crank D (diameter 33mm), which has a reverse motion linkage system with rod H2 (33mm) attached to a fixed object, and rod H (70mm) attached to a cylinder which it pushes through a path with a total travel of 35mm ( the crank moves through the path too). The lever L has a range of motion from perpendicular through to 45 degrees below the horizontal. I've chosen θ 55 degrees, because that is essentially where resistance starts, and where the pivot points of H and H2 on the crank are at horizontal.
H and H2 pivots are a=6.5mm from the perimeter of the crank, so they are at radius r=10mm.
My diagram is a bit poor as it doesn't show that H and H2 have pivots on their ends, so their angle changes throughout the motion.
I'm looking to calculate the force exerted by H2 on the bottom cylinder,
or rather the mechanical advantage throughout the range of motion.

Homework Equations


Torque
T = F * L * sin(90-theta)
Pythagoreum
r^2 +H^2 = X^2
Crank position
x = r.cos A + sqrt{H^2 - r^2\sin^2 A}
Constant
Angle A = θ + 45 degrees

The Attempt at a Solution



I can calculate the torque on the crank with a perpendicular force of 186 N applied as pictured at θ 55 degrees,
to T= 21.34 Nm.
And at this moment the distance X perpendicular from the crank to the bottom cylinder = 70.7mm

but what is the perpendicular force applied on the cylinder by H at this moment?
How does one calculate mechanical advantage?

I've looked through slider crank materials, but it is always with regard to angular velocity,
I'm only interested in the perpendicular force applied on the bottom cylinder throughout the motion in relation to the perpendicular force applied to the Lever H.
Is it a problem of net torque? with the counter torque applied at a lever of the offset radius of H 10mm?thanks for the help
 

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  • #2
Welcome to the PF.

(BTW, what does your username mean?)

I'm having trouble visualizing your mechanism. Can you post a photo of it (use the UPLOAD button)? what is the application?
 
  • #3
The limit of travel on the pistons will be when the two rods are in one straight line. That will be a bit before the lever is 35 degrees below the horizontal.
The complete model will be complicated. Do you really need a general expression, or just the bounds on the mechanical advantage?
As the rods approach collinearity, the advantage will be unlimited (tend to infinity).
 
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  • #4
berkeman said:
Welcome to the PF.

(BTW, what does your username mean?)

I'm having trouble visualizing your mechanism. Can you post a photo of it (use the UPLOAD button)? what is the application?
I interpret the circle as a disc free to move in the track delimited by the vertical dotted lines. As the lever is depressed, the disc rotates, exerting forces on the two rods attached either side of its centre. Each rod is freely pivoted at each end.
The top piston is fixed, so all the travel is in the lower piston.
I'm guessing that the lever's highest position, "perpendicular", is upright, but that is uninteresting since there is no resistance until depressed to 55 above horizontal. In that position, the attachment points of the rods are at the same heights on the disc.

Although the max depression is described as 45 below horizontal, the max separation of the pistons will occur slightly short of 35 below horizontal.
 
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  • #5
berkeman said:
Welcome to the PF.

(BTW, what does your username mean?)

I'm having trouble visualizing your mechanism. Can you post a photo of it (use the UPLOAD button)? what is the application?

Thanks,
the nickname is a play on perfume terms, powdery scent being one category.

The application is a lever espresso machine.
I'll attach some pics

lever.jpg
piston2.jpg
piston.jpg
haruspex said:
The limit of travel on the pistons will be when the two rods are in one straight line. That will be a bit before the lever is 35 degrees below the horizontal.

Yes, that is probably right,
as you can see the Lever is actually in a v type shape, my estimations of angle were perhaps not the best.

haruspex said:
I'm guessing that the lever's highest position, "perpendicular", is upright, but that is uninteresting since there is no resistance until depressed to 55 above horizontal. In that position, the attachment points of the rods are at the same heights on the disc

Correct I meant to say vertical, although considering the v shape, a straight line from lever handle to crank centre would be at ~350 degrees at the zero point of this application. Let's say angle alpha would be more suitable to describe the lever position, so at 350 degrees it opens an inlet for water to fill the chamber and up until 35 degrees, the lever moves the piston body to close the water inlet and start pressing the water in the chamber. From this point the force applied becomes critical to the application and reaches its maximum in between 35 degrees to 100 degrees, say. From 100 degrees to ~130 degrees, the calculation is secondary in importance.

haruspex said:
I interpret the circle as a disc free to move in the track delimited by the vertical dotted lines. As the lever is depressed, the disc rotates, exerting forces on the two rods attached either side of its centre. Each rod is freely pivoted at each end.
The top piston is fixed, so all the travel is in the lower piston.

Yes, exactly!
I'm actually looking to translate vertical force applied on the lever, at any alpha angle,
into pressure applied by the piston.
Piston diameter = 38 mm

haruspex said:
The complete model will be complicated. Do you really need a general expression, or just the bounds on the mechanical advantage?
As the rods approach collinearity, the advantage will be unlimited (tend to infinity).

A general expression would be nice :)
The range of the mechanical advantage would certainly be helpful.thanks
 

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  • #6
I plotted the x displacement versus crank angle A for both rod h and rod h2 but I lost the file as the laptop crashed,
in any case it was theoretical validation of the piston travel, allowing me to see if there were a full 180 deg range of motion the full travel would be 4cm,
and that with 3.5cm travel the actual full range of rotation is ¬142 deg.
I think I'll get a protractor to measure the angle relationships more precisely.

I also tried working out some calculations as I waited for feedback.
crank crvl.png


Lever angle relationship to Crank Angle
Θ+55deg+A= 180 deg

If F is a vertical force applied on the end of rod l,
then CW torque on the crank is:
τ=F.l.sin(Θ)
at torque equilibrium
CCW torque from the upward force Fp from fixed point P along rod h to radius r, at an Angle B is:
τ1=Fp.r.sin(B)
and CCW torque from the downward force Ff from fixed point F along rod h2 to radius r, at an Angle B2 is:
τ2=Ff.r.sin(B2)

τ=τ1+τ2

Is t1=t2?
offering equal resistance to the crank turning CW?

Upward force from the piston F1=Fp.cos(C)
Downward force from Fixed point F2=Ff.cos(c2)

F.l.sin(Θ)=Fp.r.sin(B) + Ff.r.sin(B2)
F.l.sin(Θ)=F1.r.sin(B)/cos(C) + F2.r.sin(B2)/cos(C2)

Is this right?
How to proceed?

thanks
 

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  • #7
powdernose said:
Is t1=t2?
No. But what is the relationship between the vertical force components F1 and F2?

Given the various lengths that you know, what is the relationship between angles A and B?
 
  • #8
powdernose said:
If F is a vertical force applied on the end of rod l,
A human pressing on the lever would probably exert the force along the perpendicular if possible. E.g. If there is a vertical support to brace against, or some friction from the floor.
 
  • #9
haruspex said:
A human pressing on the lever would probably exert the force along the perpendicular if possible. E.g. If there is a vertical support to brace against, or some friction from the floor.

Probably, or at least closer to the perpendicular,
but I want the vertical force because I can put a scale under the device and measure the vertical force.

haruspex said:
No. But what is the relationship between the vertical force components F1 and F2?

I have no idea.
At this point I'm really looking more for answers than questions.
Is F1=F2?
Because of F2, δx displacement along x2 happens, and because the point is fixed at the end of rod h2, the displacement is actually a downward displacement of the crank.
F1 concurrently displaces piston P with δx displacement along x. This pushes the piston downward.
δx displacement along x2 = δx displacement along x.
So the Piston is moving downward by 2.δx
So F1+F2= total vertical force on the piston P?

haruspex said:
Given the various lengths that you know, what is the relationship between angles A and B?

When I had plotted x and x2 from 0 deg - 180 deg using the crank position formula,
I then used the law of sines so that
sin(B)=(sin(A).x)/h
and sin(B2)=(sin(A).x2)/h2

Yesterday I used a protractor to get as precise as possible measurement of the angle relationship,
I was close, Θ+53deg+A= 180 deg
and the complete range is 135 deg
and if the chamber is filled with water optimally, resistance starts just short of Θ=30deg

After that, instead of plotting x and x2 again,
I calculated angle C from the law of sines
sin(C)=(sin(A).r)/h
A+B+C=180 deg
So I have the angles B plotted through the range of motion.
 
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  • #10
haruspex said:
what is the relationship between the vertical force components F1 and F2?
powdernose said:
I have no idea.
Assume the disc is not accelerating. What are the forces on it?
powdernose said:
F1+F2= total vertical force on the piston P?
No. Compressions in series do not add.
powdernose said:
sin(B)=(sin(A).x)/h
Yes, but x varies too. Can you find a relationship between A and B which only brings in constants?
 
  • #11
haruspex said:
Assume the disc is not accelerating. What are the forces on it?

By the disk do you mean the crank? Or the piston disk?
I suppose it doesn't matter as they move downwards in unison.

Assuming equilibrium,
F=m. a
Acceleration =0 so net Force =0

The Lever rod l only creates torque,
So vertical component F1 must equal vertical component F2, for the crank to be stationary or to move with a constant velocity down the path (which it doesn't, mostly because the resistance provided the other side of the column of water is variable)

Shouldn't the horizontal components of the forces on the h and h2 rod also be equal?
Or are those compression forces?

haruspex said:
No. Compressions in series do not add.

I don't think I understand this, do you have a link I can read up on?

haruspex said:
Yes, but x varies too. Can you find a relationship between A and B which only brings in constants?

Well A, B and x are all variables. As is Θ.
Do you mean to solve to a single variable?

A+53+Θ=180
Α=127-Θ

A+B+C=180
B=180-A-C
B=53+Θ-C

From the law of sines as mentioned above
sin(C)=(sin(A).r)/h
So, arcsin[sin(C)] =arcsin[(sin(127-Θ).r)/h]
C=arcsin[(sin(127-Θ).r)/h]

B=53+Θ-arcsin[(sin(127-Θ).r)/h]

Similarly
B2=53+Θ-arcsin[(sin(127-Θ).r)/h2]
 
  • #12
powdernose said:
vertical component F1 must equal vertical component F2
Yes.
powdernose said:
Shouldn't the horizontal components of the forces on the h and h2 rod also be equal?
No. The crank is constrained to move vertically, so there are horizontal forces from the sleeve that constrains it.
powdernose said:
B=53+Θ-arcsin[(sin(127-Θ).r)/h]
Yes.
 
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  • #14
Thanks for the link.

haruspex said:
No. The crank is constrained to move vertically, so there are horizontal forces from the sleeve that constrains it.

Actually, the sleeve is never in contact with the crank, my first diagram is inaccurate in that aspect,
but I'd said the crank has a diameter of 3.3cm, while the cylinder is 3.8cm in diameter.
How does that change things?

So to continue:
τ=τ1+τ2
F.l.sin(Θ)=Fp.r.sin(B) + Ff.r.sin(B2)
Upward force from the piston F1=Fp.cos(C)
Downward force from Fixed point F2=Ff.cos(c2)
and F1=F2=Force applied Fa

F.l.sin(Θ)=Fa.r.sin(B)/cos(C) + Fa.r.sin(B2)/cos(C2)
F.l.sin(Θ)=Fa.r(sin(B)/cos(C) + sin(B2)/cos(C2))

M.Advantage =Fa/F=[l.sin(Θ)]/[r.(sin(B)/cos(C) + sin(B2)/cos(C2))]

C=arcsin[(sin(127-Θ).r)/h]
C2=arcsin[(sin(127-Θ).r)/h2]
B=53+Θ-arcsin[(sin(127-Θ).r)/h]
B2=53+Θ-arcsin[(sin(127-Θ).r)/h2]

M.A=Fa/F=[l.sin(Θ)]/[r.(sin(53+Θ-arcsin[(sin(127-Θ).r)/h])/cos(arcsin[(sin(127-Θ).r)/h])+sin(53+Θ-arcsin[(sin(127-Θ).r)/h2])/cos(arcsin[(sin(127-Θ).r)/h2])]And just to check, as I've reached this far,
once the force is applied to the piston disk of D=3.8cm,
a pressure f/area is exerted on the column of water in the cylinder
water cylinder.png


At the bottom of the cylinder the diameter widens to D= 4.4cm,
but the pressure remains the same yes?
and F1/A1=F2/A2?

thanks
 

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  • #15
First, a correction
powdernose said:
So vertical component F1 must equal vertical component F2,
No, there is a linear force from the lever as well.
powdernose said:
the sleeve is never in contact with the crank
Then what stops the crank moving sideways? Wouldn't it require that the force applied to the lever is at just the right angle?
 
  • #16
haruspex said:
First, a correction
No, there is a linear force from the lever as well.

Okay, I think I see what you mean.
If I don't assume optimal operation with the force applied perpendicular to the lever, then I will have a linear force along the lever towards the crank,
and this force will have a vertical component,
which should be F3=cos^2(Θ).F
And so it would follow that
F1=F2+F3=Force applied Fa

haruspex said:
Then what stops the crank moving sideways? Wouldn't it require that the force applied to the lever is at just the right angle?

Good question.
I looked at the machine again and going back to post #5
in the first picture
powdernose said:
I'll attach some pics

View attachment 214259

one can see that the rod fits into a vertical slot, so its motion is constricted to the vertical, hence a horizontal 'sleeve' for the crankAny thoughts on the pressure?
powdernose said:
At the bottom of the cylinder the diameter widens to D= 4.4cm,
but the pressure remains the same yes?
and F1/A1=F2/A2?
 
  • #17
powdernose said:
If I don't assume optimal operation with the force applied perpendicular to the lever,
No, there will be a vertical linear contribution from the applied force unless it is horizontal.
powdernose said:
one can see that the rod fits into a vertical slot, so its motion is constricted to the vertical, hence a horizontal 'sleeve' for the crank
Not sure what you are saying there. It looks to me that where the lever curves around to form the crank axle, and enters the cylinder, it passes through a pair of vertical plates. I would guess these plates can slide vertically in the wall of the cylinder. If so, that effectively constrains the axle against horizontal movement. Is that what you mean?

Re the pressure, as I remarked, this becomes theoretically unlimited as the pistons reach maximum separation. I.e. the mechanical advantage becomes infinite.

This appears to be a very practical problem, so I assume you after some practical result, not necessarily requiring a perfectly correct equation for the mechanical advantage as a function of theta. If so, we might be able to take a short cut.
 
  • #18
haruspex said:
No, there will be a vertical linear contribution from the applied force unless it is horizontal.

Okay,
so the crank disk needs to apply an equal and opposite upward force on the other end of the lever,
and that force pairs with a downward force on the crank disk?
And in the case that the force applied on the lever is always vertical, then the force will =F
I guess I kept thinking the torque was the opposite reaction, when one needs to balance linear forces and torque.

τ=τ1+τ2
F.l.sin(Θ)=Fp.r.sin(B) + Ff.r.sin(B2)
Upward force from the piston F1=Fp.cos(C)
Downward force from Fixed point F2=Ff.cos(c2)
Downward force from lever link = F
and F1=F2+F?

Thinking about it more, I wondered if one should consider the mass of the lever and the gravitational pull, and the resulting moment from such.
I guess one would have to for a correct model.
And the crank disk mass and grav. pull too? Or is that always in effect countered by the anchoring fixed point?

haruspex said:
Not sure what you are saying there. It looks to me that where the lever curves around to form the crank axle, and enters the cylinder, it passes through a pair of vertical plates. I would guess these plates can slide vertically in the wall of the cylinder. If so, that effectively constrains the axle against horizontal movement. Is that what you mean?

The cylinder is slotted. The vertical slots are cut into the cylinder. But yes, that is the effect.

haruspex said:
Re the pressure, as I remarked, this becomes theoretically unlimited as the pistons reach maximum separation. I.e. the mechanical advantage becomes infinite.

I recall you mentioning an unlimited advantage at co-linearity, but maximum separation doesn't occur ever.

I was asking more about the pressure in isolation,
forgetting a crank system,
say once a simple constant vertical force is applied to the piston disk of D=3.8cm,
a pressure f/area is exerted on the column of water in the cylinder
At the bottom of the cylinder the diameter widens to D= 4.4cm,
but the pressure remains the same yes?
and F1/A1=F2/A2?

haruspex said:
This appears to be a very practical problem, so I assume you after some practical result, not necessarily requiring a perfectly correct equation for the mechanical advantage as a function of theta. If so, we might be able to take a short cut.

Well yes, it is a practical problem, and a practical result would be good.
Although I'd say I've put in work towards the theoretical.
 
  • #19
powdernose said:
τ=τ1+τ2
F.l.sin(Θ)=Fp.r.sin(B) + Ff.r.sin(B2)
Upward force from the piston F1=Fp.cos(C)
Downward force from Fixed point F2=Ff.cos(c2)
Downward force from lever link = F
and F1=F2+F?
That all looks right.
powdernose said:
one should consider the mass of the lever and the gravitational pull, and the resulting moment from such.
The weight, yes, but you can just add this to the linear force and torque from F as appropriate.
powdernose said:
And the crank disk mass and grav. pull too?
The weight, yes.
powdernose said:
maximum separation doesn't occur ever.
Recall I wrote in post #3:
haruspex said:
The limit of travel on the pistons will be when the two rods are in one straight line. That will be a bit before the lever is 35 degrees below the horizontal.
Your diagram shows the lever at 55 above horizontal and the rod attachment points on the same level. You also said that the low point of the lever is 45 degrees below horizontal, so that is a rotation by 100 degrees from the diagram position.
After rotating by 90 degrees the attachments points will be in a vertical line. That is beyond the point where the distance between the pistons is at a maximum.
So is the 45 degrees below statement wrong?
(Btw, the rods are never in a straight line because the crank disk is nearer to one piston than the other.)
powdernose said:
say once a simple constant vertical force is applied to the piston disk of D=3.8cm,
a pressure f/area is exerted on the column of water in the cylinder
At the bottom of the cylinder the diameter widens to D= 4.4cm,
but the pressure remains the same yes?
and F1/A1=F2/A2?
Yes.
 
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  • #20
haruspex said:
Recall I wrote in post #3:
That is beyond the point where the distance between the pistons is at a maximum.
So is the 45 degrees below statement wrong?
(Btw, the rods are never in a straight line because the crank disk is nearer to one piston than the other.)

Yes, I believe I said you were right in post #5,
While my estimation of 55 deg was close, because I was looking at the effort arm,
when looking at the bottom position I initially, absent minded, looked at the shorter side of the V bar.
I did use a protractor later to get better estimations,
for which A+53+Θ=180
and estimated the full range of motion for Θ to be from -18 deg down to 117 deg.
Co-linearity would occur at Θ=127.

haruspex said:
That all looks right.

Okay,
then, ignoring the weight contributions of the mechanical parts for now, we have:
F.l.sin(Θ)=F1.r.sin(B)/cos(C) + (F1-F).r.sin(B2)/cos(C2)
F.l.sin(Θ)=F1.r.sin(B)/cos(C) + F1.r.sin(B2)/cos(C2) -F.r.sin(B2)/cos(C2)
F.l.sin(Θ) + F.r.sin(B2)/cos(C2)=F1.r.sin(B)/cos(C) + F1.r.sin(B2)/cos(C2)
F(l.sin(Θ)+r.sin(B2)/cos(C2))=F1.r(sin(B)/cos(C) + sin(B2)/cos(C2))
F1/F=(l.sin(Θ)+r.sin(B2)/cos(C2))/r.(sin(B)/cos(C) + sin(B2)/cos(C2))

C=arcsin[(sin(127-Θ).r)/h]
C2=arcsin[(sin(127-Θ).r)/h2]
B=53+Θ-arcsin[(sin(127-Θ).r)/h]
B2=53+Θ-arcsin[(sin(127-Θ).r)/h2]

M.A Vertical=Fa/F=[l.sin(Θ)+r.sin(53+Θ-arcsin[(sin(127-Θ).r)/h2])/cos(arcsin[(sin(127-Θ).r)/h2])]/[r.(sin(53+Θ-arcsin[(sin(127-Θ).r)/h])/cos(arcsin[(sin(127-Θ).r)/h])+sin(53+Θ-arcsin[(sin(127-Θ).r)/h2])/cos(arcsin[(sin(127-Θ).r)/h2])]

And for a constant perpendicular force applied on the lever, similarly it follows that:
F.l=F1.r.sin(B)/cos(C) + (F1-Fsin(Θ)).r.sin(B2)/cos(C2)
F1/F=(l+r.sin(Θ).sin(B2)/cos(C2))/r.(sin(B)/cos(C) + sin(B2)/cos(C2))

M.A Perpendicular=Fa/F=[l+r.sin(Θ).sin(53+Θ-arcsin[(sin(127-Θ).r)/h2])/cos(arcsin[(sin(127-Θ).r)/h2])]/[r.(sin(53+Θ-arcsin[(sin(127-Θ).r)/h])/cos(arcsin[(sin(127-Θ).r)/h])+sin(53+Θ-arcsin[(sin(127-Θ).r)/h2])/cos(arcsin[(sin(127-Θ).r)/h2])]

I then plotted the M.A through the range of motion

upload_2017-11-15_2-42-29.png


does that seem right?

haruspex said:
The weight, yes, but you can just add this to the linear force and torque from F as appropriate.

But then I lose the M.A as a function of Theta, right?
because by adding just the weight of the lever F4:
F.l.sin(Θ) + F4.l/2=F1.r.sin(B)/cos(C) + (F1-F-F4).r.sin(B2)/cos(C2)
->
F1=(F(l.sin(Θ)+r.sin(B2)/cos(C2)) + F4.(r.sin(B2)/cos(C2) - l/2))/r.(sin(B)/cos(C) + sin(B2)/cos(C2))
So there is a Theta dependent constant from F4 always offsetting the Mechanical advantage?

I ran some calculations across a small theta range with varying forces applied to the lever at a range of forces suitable to the application and found that the 2.18N weight of the lever reduces the Mechanical advantage (in the vertical scenario) by 1-3%.
I've not estimated with the crank weight because I'm not going to dismantle the mechanism to weigh it.
 

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  • #21
I just realized I had been misled by your original diagram when I wrote this:
haruspex said:
After rotating by 90 degrees the attachments points will be in a vertical line. That is beyond the point where the distance between the pistons is at a maximum.
You had shown the rods as verical there, so the ponts of attachment to the pistons were not central. Seeing the correction in post #6, I am now happy that the rod colinearity coincides with verticality.
powdernose said:
down to 117 deg.
Yes, but that's 27 below horizontal, a long way short of 45. It's the 45 that implied the pistons would reach, and return from, maximum separation.
powdernose said:
Co-linearity would occur at Θ=127.
Yes.
powdernose said:
F.l.sin(Θ) + F4.l/2
Doesn't the sin(θ) factor apply to F4 there?
 
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  • #22
haruspex said:
I just realized I had been misled by your original diagram when I wrote this:
You had shown the rods as verical there, so the ponts of attachment to the pistons were not central. Seeing the correction in post #6, I am now happy that the rod colinearity coincides with verticality.

Yes, I'm sorry about the initial mistake with the angles,
I tried to edit the post when I sorted the angles out, but I couldn't edit the post by then.

haruspex said:
Doesn't the sin(θ) factor apply to F4 there?

Yes that is correct.
F.l.sin(Θ) + F4.sin(Θ).l/2=F1.r.sin(B)/cos(C) + (F1-F-F4).r.sin(B2)/cos(C2)
->
F1=(F(l.sin(Θ)+r.sin(B2)/cos(C2)) + F4.(r.sin(B2)/cos(C2) - sin(Θ).l/2))/r.(sin(B)/cos(C) + sin(B2)/cos(C2))

And if I were to add the weight contribution of the disk it would be?
F.l.sin(Θ) + F4.sin(Θ).l/2=F1.r.sin(B)/cos(C) + (F1-F-F4-F5).r.sin(B2)/cos(C2)
->
F1=(F(l.sin(Θ)+r.sin(B2)/cos(C2)) + F4.(r.sin(B2)/cos(C2) - sin(Θ).l/2) + F5.r.sin(B2)/cos(C2))/r.(sin(B)/cos(C) + sin(B2)/cos(C2))

I tried to weigh the piston mechanism,
and estimated the disk to weigh 0.64N,
with this and by adding the sin(θ) factor, reduction of the mechanical advantage is under 2% at the lighter end of force applications to the mechanism,
and under 1% at greater forces.
The fixed point's weight is held by an anchor on the cylinder,
the piston weight is held by some of the force from the water bed under pressure? So it is a minimal constant subtracted from F1?
That leaves the weight of rod h and h2, which would complicate the torque function,
and the friction between the cylinder and the piston?
To be complete with all factors that is.But if my plot of the mechanical advantage is correct and the effect of adding the weight contributions is under 1%,
I can be satisfied that this is solved.

I did have a question I thought of,
going back to the piston pressure and forgetting the mechanism providing the force:
water cylinder air.png


If air is caught in the cylinder at the top,
how does that affect the pressure P2?
Is part of the pressure P1 dissipated in compressing the air bubble?
I tried searching for some help with this type of situation but coulnd't find anything.

thanks
 

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  • #23
powdernose said:
Is part of the pressure P1 dissipated in compressing the air bubble?
It depends exactly what you mean by that. For a given movement of the arm, less force is required, so less pressure is created. But for a given force applied, the pressure is the same - it just requires the arm to be moved further.
 
  • #24
haruspex said:
It depends exactly what you mean by that. For a given movement of the arm, less force is required, so less pressure is created. But for a given force applied, the pressure is the same - it just requires the arm to be moved further.

I'm not sure I follow.

I think it would be easier for me to understand taking the cylinder pressure in isolation.
water cylinder air.png


F1 applied to the piston (3.8cm) on top of a column of water, applies pressure P1,
For equilibrium at the bottom (4.4cm) must have a pressure P2
P1=P2 in case1 and case3 only?
The pressure at the top must compress the air fully before equilibrium can be achieved?

I don't understand how it works with water and air mixed in the cylinder.

powdernose said:
I tried to weigh the piston mechanism,
and estimated the disk to weigh 0.64N,
with this and by adding the sin(θ) factor, reduction of the mechanical advantage is under 2% at the lighter end of force applications to the mechanism,
and under 1% at greater forces.
The fixed point's weight is held by an anchor on the cylinder,
the piston weight is held by some of the force from the water bed under pressure? So it is a minimal constant subtracted from F1?
That leaves the weight of rod h and h2, which would complicate the torque function,
and the friction between the cylinder and the piston?
To be complete with all factors that is.But if my plot of the mechanical advantage is correct and the effect of adding the weight contributions is under 1%,
I can be satisfied that this is solved.

Is my line of thinking correct here?

And is my theoretical model for the Mechanical advantage good enough?

thanks
 

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  • #25
powdernose said:
P1=P2 in case1 and case3 only?
No, they are equal in all cases.
The difference is that with air present the piston will have to move further to achieve the same pressure.
 
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  • #26
haruspex said:
No, they are equal in all cases.
The difference is that with air present the piston will have to move further to achieve the same pressure.

Okay.

I assume the mechanical advantage problem has been adequately solved,
thanks for the guidance
 
  • #27
powdernose said:
Okay.

I assume the mechanical advantage problem has been adequately solved,
thanks for the guidance
Yes, I think it is ok.
 
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Related to Mechanical Advantage in a crank system

1. What is mechanical advantage in a crank system?

Mechanical advantage refers to the ratio of the output force to the input force in a crank system. It measures how much the system amplifies or multiplies the input force to produce a greater output force.

2. How is mechanical advantage calculated in a crank system?

Mechanical advantage can be calculated by dividing the distance from the fulcrum (pivot point) to the point of application of the input force by the distance from the fulcrum to the point of application of the output force.

3. What are the advantages of using a crank system?

Using a crank system can provide a mechanical advantage, making it easier to lift or move heavy objects. It also allows for the transfer of energy from one part of a machine to another, allowing for more efficient and precise movements.

4. What factors affect the mechanical advantage in a crank system?

The length of the crank, the size of the gears, and the angle at which the force is applied all affect the mechanical advantage in a crank system. A longer crank or larger gears can increase the mechanical advantage, while a smaller angle of force application can decrease it.

5. Can a crank system have a mechanical disadvantage?

Yes, a crank system can have a mechanical disadvantage if the input force is greater than the output force. This can occur if the angle of force application is too small or if there is friction present in the system, causing energy loss.

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