GreenSteam engine

[Jonathan]

Ken, C is the closest, there's a bent rod just to the left of it. The thing its pointing to is a chunk of metal!
If you're going to use solenoids, then you might as well throw out the whole thing and use a DC motor.

[rlortie]

Ken,

If you are still confused of the bent rod, I have edited the following attachment highlighting said shaft. note that it acts as an axle but is not centered with flywheel. Something to think about in your Bessler designs.

If one looks closely you will see that there is no reason for the shaft to flex. It could be a perminent bent supported in the flywheel by a pivot joint.

Ralph

[Michael]

Michael,

Where are you heading with the rack and pinion gears depicted above. is this implying something to Mitch personally.



Ralph

>There's still a loss of efficiancy though. It may not have much friction but everytime the angle from the rotary part to the piston starts to change ie: changes from pulling at a 90 degree angle relative to the rotary parts axel, efficiancy goes down. It's the same with crankshafts. The thing to do would be to keep the 90 degree angle. A straightline sawtooth gear comes to mind, or something better along this type of thinking. Simple. Can be made smaller. I wonder why it hasn't been developed and refined before.

[Michael]

image 2

[Michael]

Reason?

When the angle is at 90 degrees the rotary part is recieving the full amount of force over the distance traveled from the piston movement. When the angle changes the piston has to move faster because the rotary part is starting to move faster but there is also a loss of leverage in the process. Result, a lack of efficiancy.

One possible solution?

[rlortie]

Michael,

You ask why hasn't it been developed and refined before?

I hate to be the one to tell you this but just about every auto on the road is now equipped with rack and pinion steering. Where the pinion is usually the input end. You have reversed it to the rack.

A ring gear around a planetary gear drive is a rack bent into a circle.

I agree with you that when attached to a ratchet drive and a sufficient flywheel it would transfer a high percentage of force for the input.

By the way, do not forget that a steam engine developes maximum torque at zero rpm when given a full head of steam.

Ralph

[Michael]

Ralph, respectively your first comment has taken what I wrote out of context. I'm not talking about steering or the development of rack and pinion, I am talking about engines and I am showing that a rack and pinion system or a better developed but like system is more efficient than a crank or it's equivilent (like this engine uses) system, as you admit. Why hasn't this been developed towards all car engines, that's the question.

Mike

[Jonathan]

Michael, I think they don't use rack and pinion instead of a crankshaft because the former would be so noisy as to sell poorly, not to mention the loss of efficiency due to the noise.
Ralph, if the bent rod were permanently bent, then the thing attached between it and the flywheel would have little incentive to stay horizontal as it went around. (This 'thing' is also the thing that the piston rods and valve rod connect to; it's labeled C in one of Ken's attachments). The important idea in this design is that the bar can bent but not twist. The 'thing' is attached to a flexible but non-twisting bar, so it can travel in a circle and maintain a constant orientation.

[Michael]

>Michael, I think they don't use rack and pinion instead of a crankshaft because the former would be so noisy as to sell poorly, not to mention the loss of efficiency due to the noise.


I know Jonathan, :) , that's why I said developed and refined. Something along those lines. As for rack and pinion, I don't think there will be that much loss do to noise if the gears are made to mesh well but there will still be some. I can see a flexible rod made from linkage where the parts pivot doing a good substitute job, but then there's an added possibility of breakage.

[Jonathan]

How about this then? Fourteen links, for two pistons, 180° out of phase. The applied force is always in line with the velocity of the point of application, so it should be quite efficient, with light but strong parts, and good lubrication. (The points of application are the bottom two corners of the picture; the pistons and cylinders would form a right angle V).
EDIT PS. I am aware that this only produces reciprocating rotary motion, but there are ways to convert it to a normal rotary output.

[ken_behrendt]

Well, I reread this entire thread and revisited the link to the steam engine and I still haven't the faintest idea of how this gadget is supposed to operate!

Jonathan wrote:

Ken, C is the closest, there's a bent rod just to the left of it. The thing its pointing to is a chunk of metal!

That bent rod to the left of the part I labeled C looks to me like its made of rigid metal. Maybe the part I labeled B is the flexible elastic piece? It's white and looks like it's plastic.

In any event, even if I knew which piece was the flexible rod, I still do not understand how this device converts reciprocating motion into rotary motion. I tried viewing the video at the side, but the motion is a blur and I can not see what's going on clearly.

Anyway, I agree...I'd skip the steam engine and just use a simple DC motor where all I have to worry about is bearing failure and worn out brushes...


ken

[Michael]

If you think about it 90 degrees is the greatest amount of force, 0 degrees is the least. That would mean the average force delivered to a typical crankshaft would be about half of what is possible. Imagine, your current car being able to deliver twice the amount of power, or getting twice the amount of travel for the same amount of gas.

Michael

[Jonathan]

Ken, it looks like rigid metal, but it's not. It's like those springy things (attached to walls behind doors, near the floor) that keep the door knob from hitting the wall.

[jim_mich]

If you think about it 90 degrees is the greatest amount of force, 0 degrees is the least. That would mean the average force delivered to a typical crankshaft would be about half of what is possible.

No Michael, it doesn't quite work that way. You get less torque but you get more rotational travel. Say you have an eccentric crankshaft radius of 1 unit. This means the piston will move 2 units. Compare it to a rack, the rack would move 2 units. But the rotation of the crankshaft at the eccentric radius would move Pi (3.14159) units. When you measure the average force on the crankshaft and multiply times the rotational distance moved, then the loss if actually quite small. An imaginary rod length of infinity pushing an eccentric radius of 1.00 would equal a rack driving a gear with a 1.00 radius. As the rod length becomes shorter there is a small loss which is not significant until the rod to eccentric ratio becomes small. Of course with a rack and pinion the piston velocity is constant while with a crankshaft the velocity peaks at mid-stroke. This gives the piston time to accelerate and decelerate which would be difficult with a rack and pinion.

Code: Select all

infinite rod ratio = 0.6366197 * Pi = 2.0000 units = No Loss

1:10 rod ratio = 0.63449346 * Pi = 1.9933 units = -0.3% Loss
1:6  rod ratio = 0.63069202 * Pi = 1.9814 units = -0.9% Loss
1:3  rod ratio = 0.61249070 * Pi = 1.9242 units = -3.6% Loss
1:2  rod ratio = 0.58058426 * Pi = 1.8240 units = -8.8% Loss

Basic Program...
DEFDBL A-Z       'All variables double precision

Pi = ATN(1#) * 4
d2r = Pi / 180   'degree to radian conversion factor

R1 = 1  ' CrankShaft eccentric radius
R2 = 3  ' Piston Rod Length. Change this for different ratios.

T = 0   'Zero total
K = 0   'Zero count
S = .01 'Step angle

FOR d = -90 + S / 2 TO 90 STEP S
 a = d * d2r      'angle as radians
 X1 = COS(a) * R1 'CrankShaft Eccentric location
 Y1 = SIN(a) * R1

 Y2 = SQR(R2 ^ 2 - X1 ^ 2) 'Vertical distance eccentric to piston pin.

 Y3 = Y1 + Y2  'Piston pin location
 X3 = 0        

 A3 = ATN(X1 / Y2) 'Angle piston rod tilts to one side.
 R3 = SIN(A3) * Y3 'Radial distance of force vector from center.

 F = R3 / R1  'Torque Force

 T = T + F  'Total sum of torque
 K = K + 1  'Count
NEXT
PRINT
PRINT T
PRINT "/"; K
PRINT "-----------"
PRINT T / K; " average torque force"
PRINT T / K * Pi; " torque distance"
PRINT R1; ":"; R2 " rod ratio"
PRINT

[Michael]

Your right Jim, because I was loosely generalizing. The losses are still fairly substantial though. I'll explain a different way and draw you a picture.
Here are points I agree with.

>You get less torque but you get more rotational travel.

Yes, check previous messages.

>Say you have an eccentric crankshaft radius of 1 unit. This means the piston will move 2 units.

Yes

>Compare it to a rack, the rack would move 2 units.

Yes.


>But the rotation of the crankshaft at the eccentric radius would move Pi (3.14159) units.

Yes


>This gives the piston time to accelerate and decelerate which would be difficult with a rack and pinion.

It's true that as a gas expands the piston will loose power. This is a consideration when making something workable but doesn't take away from the benifit of using a like rack and pinion system. One can always make corrections for that.

>An imaginary rod length of infinity pushing an eccentric radius of 1.00 would equal a rack driving a gear with a 1.00 radius. As the rod length becomes shorter there is a small loss which is not significant until the rod to eccentric ratio becomes small.

For what I am talking about this doesn't apply.