Naked Science Forum
Non Life Sciences => Technology => Topic started by: xengineguy on 03/08/2008 18:51:22
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Hi everyone,for some time I have been asking a question about how to compare two ICE engines. In most forums I get no response,so lets try here.
I'm looking for a way to compare two engines,Same bore,same stroke,same cylinder pressures. The difference is the first engine is a conventional crankshaft engine,and the second has a very different (crank-less) system.
EXAMPLE: First engine,bore 3" stroke 3", max leverage @ 90degrees ATDC is 1.5",peek cylinder pressure is 4000
at 20 degrees ATDC. Cylinder pressure drops to 400 @ 90 ATDC.
Second engine same as the first accept:Leverage @ 90 ATDC is 12", At TDC is 10.5",at BDC is 13.5" Is this enough information????? Not looking for hosepower just a "work"???? comparison.
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Hello xengineguy
I don't understand the second specification. The length of crank has to relate to the stroke of the engine - small end bearing goes in a circle- so where do you get your 12" leverage from? How can it have anything other than 1.5"?
Perhaps I have not understood your figures. What exactly do you mean?
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- small end bearing goes in a circle-
Does it? The small end is the bearing where the piston head fixes to the shaft (not sure of the technical term). That means the bearing will only oscillate through a relatively small angle - the longer the stroke, the greater the angle; but it can never go in a circle. The big end is the bearing that goes in a circle as that is where the piston assembly is fixed to the crankshaft.
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you're right - but the 1.5" bit still applies. I can't imagine what the figures mean.
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Hi again,
As stated the second engine is a "crankless" design. The piston is no longer "tied" to a crank. Max leverage is no longer half the stroke. The leverage is never reduced to 0" at TDC and BDC like a crank engine. The figures provided are correct for comparison. The cylinder pressures can be adjusted to whatever value as long
as they are the same for both engines. I have a running prototype,and willing to show it,but it seems to be a distraction when I try to get an answer to this question...Thank you for your comments!!!!!
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Sounds a bit like a 'beam engine'. OK if you want an oscillating output.
Is your basic question about the way the volume changes with time and how that affects the efficiency / power output?
A rotary engine fixes a sinusoidal oscillation; your engine could be anything you wanted - but how practical would that be?
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Not a beam engine,more like a rotary. Could you explain sinusoidal oscillation the negative effects as applied to an ICE.
To make things easier to picture go to YOUTUBE/marckel rotary piston engine. Please understand this is a proof of concept prototype. I know many things would be changed on a production engine!! This is the "type" of engine I describe.
I need a simple way to compare the leverage of this to a standard engine,if its better ok if not ok. Some kind of formula that fits the info we currently have.
Thank you..
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To clarify my sinusoid statement ; A 'normal' rotary ICE has the piston coupled to a crankshaft. The crankshaft goes in a circle and the big end, at least, describes a sine wave in the direction of the piston (up/down). If the con rod is long, then the piston also moves up and down with a sinusoidal motion. With short con rods, the piston motion is not exactly sinusoidal because the con rod rocks a long way from side to side.
I looked at several links and it seems the engine is a bit like the old rotary aero engine from some first world war aircraft.
I'm not sure where the 12" leverage comes from. I'd need to see a good drawing of the system to see what does what.
There are a number of difficulties associated with these alternative arrangements - lots of rotating mass (high moment of inertia).
But it looks interesting.
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The 12" comes from the piont the roller touches the outer lobe (in this case on the power ramp) measured to the center of rotation. In the question consider the ramp angle to be 45 degrees or (one to one). Seems to make things a little clearer.
So far the results seem too good...(with my calculations)???
My method: Start with equal cylinder force of 6000#,reduce that force by 375# per .125" of leverage increase.For my problem I used 6000#@ .125" for engine 1 and 6000#@ 14.125" for engine 2. The second set is 5625@ .250" for #1 and 5625@ 14.250" for #2.Force times length to get torque. Total stroke for this problem is 4".
Then I took the torque values times 1000,(rpm) divided by 5252 for a horsepower number.
The horsepower number for engine #2 was adjusted to reflect the difference in movement. 9:1 is the difference. The horsepower number for engine #2 was divided by 9.
After comparing the results I get an average horsepower increase of 2.13 times.Or over 100%!!!! Need to find out if this is correct asap.
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I think the geometry of the setup is not what you think it is; I have not seen a proper, functional, diagram of the engine but I don't see how you can get the swept volume with the same piston area and still have the 'leverage' you refer to. Do you have a simple schematic that you could post? Simple ideas like the method of moments seem to be violated by your brief description.
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You may be correct,I tried to post some dxf. cad files to help,no luck. Will work on it. I hate to ask,could you look at the prototype again? Asking specific questions about it might clear things up a little more?
The leverage I talk about is the piont the roller contacts the stator, compared to the axis of the rotating assembly. NOTE: The prototype has 1.125 stroke with apx. 6" (leverage)? Hope this helps..Make sure the engine on youtube is spelled Marckel. There is another spelled markel? and its a two cycle rotary like a gnome engine.
Thanks again
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I don't really feel inclined to look much further at this idea if the only material available about it is a Utube movie.
Using Utube for information is no better than saying "A bloke in a pub told me".
If there aren't any proper diagrams and real data available then the system is suspect. It sounds too good to be true and, without some more information, I have to feel that it may not be true.
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Looks like I wasted my time and yours.The Question was how to compare, NOT do you believe...I do have simple cad drawings but was unable to get them to post? I didn't want to even refer you to youtube but we were off track. I don't have unlimited funds,so I ask the Question on your forum. If you cant accept my description thats OK.
Whats the harm in saying "what if" and just help answer the question????
THANKS anyway
Mike
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Sorry if I sounded too skeptical but your question seemed (still does) to have an ambiguity. It seemed to violate a basic bit of engineering. If it doesn't, then fair enough but, even a 'what if' question has to have some solid basis if it is to be answerable. My problem is that your verbal description seems to imply a greater radius without a compensatory reduction in force -to keep the moments right - or a change of stroke without a reduction in piston area - to keep the work done the same.
Have you tried a screenshot of your cad files? A JPEG will post fine as long as it isn't too big. Without it I have no idea what turns what so i really can't make a comment.
This is not a grumpy reply- it is a request for info in order to proceed with the comparison you wanted. I can only compare two things if I can believe they both work.
You will appreciate that there are some ideas posted on this forum which just don't work.
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Posting pictures... [ Invalid Attachment ]
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Sorry - just a grey blob so far.
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Wow that was big! [ Invalid Attachment ]
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Will try again. [ Invalid Attachment ]
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Try again.. [ Invalid Attachment ]
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Ok one more.. [ Invalid Attachment ] [ Invalid Attachment ]
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it's not enough to design a new engine. You also have to design a a dyno to test it maybe?
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Without going into too much detail I think I have identified my problem. The 'leverage' you refer to seems very optimistic. To find the actual moment or turning effect you multiply the force by the 'perpendicular distance' of the force vector from the point where the force acts to the centre (see below). The large cam which the push rods acts on produces a force which is directed in a direction which is more towards the pivot than on a tangent. This means that the perpendicular distance is nothing like the 12" radius you have suggested. The effective radius will be more like the crank of a conventional engine. The stroke length is indicated bu the slots at the end of the pushrods.
The stroke length seems to be adjusted by the pitch of the cam and the profile of the cam can be tailored to 'thermodynamic' requirements whereas a normal crank engine imposes a sinusoidal motion.
Producing an output and getting fuel and exhaust transfered seem to be a serious problem, though.
The science bit:
[diagram=359_0]
The lever, length L has only the effect of a lever length D if the force is along the line F. (No idea why the diagram seems to have been distorted whilst posting it.) The moment is FD not FL.
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I understand,let me clear up a few things. The prototype has different dimensions than what I described in the question. The prototype would have more like 6" "Leverage". In the question I also said consider the ramp angle (cam) to be 45 degrees. With that said how do I find the effective radius? Is it 50% of the physical radius?
As far as producing a power output there is a center shaft turning,no problem.
Fuel and exhaust minor problems if the design is more efficient..
Thanks again
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OK, well the effective radius, for calculating the moment, will be the perpendicular distance between the pivot and the normal to the cam, wherever the roller happens to be. I think that is correct; it assumes that the push rod roller is free- which means that the force must be at right angles to the cam surface.
If you carefully draw a set of normals to the cam surface they will produce an envelope of the curve of how the force acts as the cam rotates. At some point, the curve will be furthest from the pivot and that will be your maximum leverage. To find the actual moment and the energy being transferred you would need to find the volume and pressure inside the cylinder at all these points as it varies in time and then get the moments from the cylinder force and the effective radius. Sounds a bit complicated because you'd have to do it numerically then integrate by adding up the work done. Sound a bit like hard work but Excel could help you.
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I have calculated the moments in .125 increments of both engines. And calculated the force at each moment(with descending pressures).I now have to factor the effective radius or leverage.
Lets see if I understand. Start with a circle 30" diameter. (This would be the path the roller follows when the piston is at BDC.) Add a 45 degree lobe to the inner surface of the circle. Draw a line from the center of the circle to the lobe.(This is the force the piston applies to the ramp.) The "L" line. Draw another line from the point the first line touches the lobe,(90 degrees to the lobe surface).This is the "F" line. Now draw another line from the center of the circle to line "F",this line must be 90 degrees to line "F"...This is moment or line "D".The distance from the center to line "F" is the effective radius. Is this correct?????
Thanks
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Sounds OK so far. You need to make sure that the 'stroke' is consistent with the geometry of the cam, of course; the engine design needs to be 'possible'. You will, presumably, have the same swept volume for both engines.
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This one is a crankshaft engine,is this correct? [ Invalid Attachment ]
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The bore/stroke are the same.Posting work based on your information. [ Invalid Attachment ]
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The geometry looks ok so far but I am not sure what your columns of numbers mean (I assume they are right). You have only drawn one section of the cam; The stroke will be (?) the difference between the maximum and minimum radius of the cam; the profile in between will affect how the volume changes through the cycle.
You may need to go over the whole cycle of 'your' engine and make sure that the cam radius and angles actually produce the same stroke length as the cranked engine.
Are you sure that the cam on the engine in the picture is as steep as 450? In the picture, it seems a lot less pitched.
I think that the force normal to the cam will be the pushrod force modified by a factor of Cos(angleACB). This means is gets less and less as the cam gets steeper and needs to be factored into the calculation each time. You will get more 'leverage' but less effective force. I think that you are being a bit over optimistic about the new design and that the advantages you are looking for will, in fact, 'all come out in the wash' and prove to be marginal.
I have a feeling that, as you can't get something for nothing in this life, you will find that the only advantage of this engine is that it gives you a non-sinusoidal piston motion and the right choice of motion could improve efficiency. I am not dismissing it because that, in itself, could be a significant advantage. I just don't know enough about engine cycles to have an opinion. The problem is that, with the exception of the Beam Engine, all piston engines that I know of use a crank and the PV etc. diagrams all take that into account. (Wankel Engines can also have this feature; you may find some info about how effective it can be.)
Of course, it does give you four strokes per rev, unlike the cranked engine. You could also imagine having a different number of cam lobes / cylinders. You could also allow your induction/compression phases to have a different profile to the ignition/exhaust phases (different cam profiles each side); another possible improvement.
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The columns of Numbers are A-C distance and the adjusted arm of A-B at different points on the ramp. After that the % of change.
The prototype engine is very different than the engine in the question.The ramp in the prototype is not 45 degrees.
I don't think the ramp angle makes any difference in total work,45 degrees is just easy to use in an attempt to understand. From what I see as the angle is increased to say (60 degrees)more torque is produced but less motion. So in the end the horse power is the same. 45 degrees is just 1 to 1 .
I have from the beginning been interested in the leverage differences,from the axis to the ramp surface and how it compares to the crank engine. I think it may be a wash also but would like to prove it one way or another. My original calculations seemed to good,I guess what I'm asking is what am I missing,or how do I prove it with an equation based on the difference in leverage.
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When you get down to it the work done by the piston will be set of forces times a set of distances all added together. That is as much as you can possibly get out of the engine. Levers and gears etc. will still involve forces and distances which must come out more or less the same (less friction etc.) They merely present the engine with the opportunity to work 'at its best'.
I think the cam system may well do a bit better than the crank system because you can tailor things to get the best out of the combustion process. It may be possible, for instance, to make sure that all the combustion occurs with the piston near the top of its stroke so that more of the energy can be extracted. But the practicalities of the implementation may nullify this advantage.
Have fun with your sums!
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Thank you for your time and responses.I understand more now than I did before. I still have a few questions,but thats the for another time. For now I will just build another prototype and do some dyno runs. This should give me the information needed to compare with other engines of similar displacements. THANKS again Mike