[GP125] All that you wanted to know on Aprilia RSA 125, and more, by Mr Jan Thiel and Mr Frits Overmars (PART 4)

Hi, Martin; normally I'm just a lurker here, figuring I don't know nearly enough to make comments. But your description of cylinders that are known to be prone to cracking has tempted me to offer an idea. As we all know, pistons are not round, with parallel sides to the skirts, but have tapered skirts and are relieved around the wristpin areas, the idea being that the heavier masses of material expand more, and that the piston will "grow" more nearly round and parallel as it gets hot. But what about cylinders? 2-stroke cylinders also have areas with greater or lesser masses of metal, yet normal practice is to machine the cylinders perfectly round and parallel, when cold. What shape do the cylinder walls take when hot? With this in mind, some of the guys with whom I raced outboard hydroplanes back in the 1960s and '70s would heat up their cylinders with a torch or oven before doing the finish-honing. Some of us went farther, making what later became known as "torque-plates" in automotive circles. We bolted down anything that normally attached to the block, such as exhaust stubs, that might have an effect on the shape of the cylinder when hot. On any engine that had a bridge in the exhaust, I would use a torch to put extra heat into that bridge immediately before running the hone through the cylinder. I didn't know whether that exhaust bridge was likely to bulge into the cylinder or away from it or sideways, but I figured that getting the whole works to something roughly resembling running condition before honing would result in a cylinder that was closer to round and parallel while hot. I don't care what shape it is when cold.

Is is possible that something like this practice might reduce a tendency to cracking in your engines?

To carry this thought a little further  .  .  .  

Here's a quote from Frits talking about cast iron sleeves in aluminum cylinder blocks on another site:

(Frits) "Pressing in a sleeve is a no-no to begin with. Shrinking it in is better, but not good enough. Because no matter how tight you make that interference fit, during use oil will creep between cylinder and liner, starting at the exhaust port.
If you remove an old liner, you can see the charred remains of that oil all around the exhaust zone. So do not worry about the limited thermal conductivity of an iron liner; charred oil conducts heat infinitely worse.
Been there, done that. If there's anything that makes me experienced in the two-stroke field, it's the fact that I have personally committed every imaginable error. Racing an air-cooled 500 cc two-stroke single with a liner made me the fastest man in Holland - in drawing a clutch. Not even Lucky Luke could draw quicker (end quote).

Frits, could it be that this situation derives not only from the fact that the aluminum cylinder not only expands faster than the iron sleeve as both heat up, but that both cylinder and sleeve are finish-machined COLD? Could the interface between sleeve and cylinder be a lot better if each part were finish-machined HOT? They'd still be sized for a shrink fit. But done this way, the sleeve-to-cylinder contact should be more uniform when the assembly gets hot, not perfectly so but much improved .  .  .  yes???

As a Band-Aid fix for sleeved cylinders done in the usual way, there was an aero-space company in California years ago that had developed a process for vacuum-drawing Loctite into crevices, for what purposes I don't know. But they evidently had some bikers among their staff, and as a tiny sideline business, they would take your sleeved cylinder and vacuum-draw Loctite into any voids between sleeve and cylinder, the idea being that while Loctite (I don't know which version) was not a particularly good conductor of heat, it was better than the voids (and maybe they mixed some aluminum powder in their Loctite). An additional and funny application involved crankcases for certain old British twins with crankcase castings so porous that oil would leak right THROUGH the castings and drip on the ground. Drawing Loctite into these cases sealed the porosity and stopped the leaks. (BTW, these were the days when Italian foundries seemed to be ahead of everybody else in making beautiful, fine-grained castings).

Sorry, I don't know the name of the company, or whether they do any of this today. Pending your comments, I think my notion of finish-machining the barrel and the sleeve hot should have a better result in any case. I know that iron sleeves are utterly obsolete for motorcycling, karting, etc, but with outboard racing we have a whole lake-full of water for cooling purposes .  .  .  COOL water that doesn't have to be pumped but is rammed into an intake hole in the bottom of the lower unit and distributed in any amount desired to any desired places in the engine. With this kind of coolant flow, iron sleeves still work for us, and have some other advantages.

Anyone's comments on any of this will be greatly appreciated.

Tim Ey a écrit:

I am asking because on the APC-drawings it is not this way.
(A-Port has got 28,16° on the roof and 16,45° on the floor,
B-Port has 6,82° on the roof and 7,42° on the floor,
C-Port has 37,7° on the roof and 52° on the floor which is never to be concentric with the Piston edge.)

Jan these were the angles for the APC cylinder.

What there the angles for your final evolution of the RSA in 2007?

The angles remained the same!

MessageSujet: Re: [GP125] All that you wanted to know on Aprilia RSA 125, and more, by Mr Jan Thiel and Mr Frits Overmars (PART 4)   Aujourd'hui à 10:06 Répondre en citant
The angles remained the same!

Mr Thiel,did the roof and the floor was like the apc cylinder (differant angle at the end)on the rsa , or the roof and floor with equal angle like Mr Overmars said it was more up to date !!????

That is difficult to say.
Actually there were no straight angles.
The inside and outside were just one continuous radius.

hi Frits or Jan thiel, How do you calculate angle.area of transfers?

hi grity i dont want to answer for fits or jan but i believe they have said how to calculate it atleast a couple times. although i dont recall the exact page. the best thing is to read and you can find it

brokedown a écrit:

hi grity i dont want to answer for fits or jan but i believe they have said how to calculate it atleast a couple times. although i dont recall the exact page. the best thing is to read and you can find it

Thanks Brokedown .

it is the same of a exhaust blowdown port?

For example, let us assume that the exhaust port is a simple rectangle, 40 mm wide, that it opens 1 mm further for each degree of crankshaft rotation, and that the total blowdown angle is 30°.
Then the first degree of exhaust opening will open an area of 40 mm x 1 mm = 40 mm², and this area will be open during the whole 30° of blowdown period. That first area thus has an angle.area of 40 x 30 = 1200°mm².

When the crankshaft turns 1 degree further, an additional area of 40 mm² is opened. This second area will be open during 29°, so its angle.area is 40 x 29 = 1160°mm².

Repeat this calculation for each crank degree until the end of the blowdown phase, and add all the angle.area values; that will give you the total blowdown angle.area.



-I have tried by this rule, but when I tried to calculate the angle.area of RSA transfers

-Assuming the 202º timing of exhaust port and 32º of blowdown and con rod 120mm and stroke 54,5mm, the exhaust open by 77,98º and sum by 32º blowdown = 109.98º (Transfers open degree) and 180-109.98º = 70.02º (180º = BDC) of transfer timing open.

- assuming that are 5 transfers with square shape and assuming that transfers are distanced of exhaust port by 3mm and between them 1,5mm; Pi*54 = 169.64-3*2-1.5*4=117.64mm its the sum of width transfers and appliing the rule in excel with the funcion of piston position  in each crankshaft angle i obtain 83504ºmm --> 110315ºmm a lot of diference,

whats is wrong in this method?  
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grity i dont understand your picture. you might want to try something like this. enter the stroke 54.5, rod length 120 and 1 degree increments then click calculate. it will tell you how far the piston travels for each degree of crank rotation. some of the port windows arent a simple rectangle shape however

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I have this funcion on first column, I assume a retangle shape to find a maximum angle.area possible, but is to far of reality. second column is the diference of positions the high of rectangule, 3rd column is piston position on radians, 4th column is timing in degree of each rectangule.

in 5th column I multiply area of each rectangule by duration of it.

example 107.92-107.51=0.409mm high of rectangle

0.409mm*117.64mm(width of retangle) * 70.02º =3371.543ºmm
0.404mm*117.64mm*69.02º=3287ºmm

ultil 0 degrees, and the sum = 83504.01ºmm

m4grity a écrit:

I have this funcion on first column, I assume a retangle shape to find a maximum angle.area possible, but is to far of reality. second column is the diference of positions the high of rectangule, 3rd column is piston position on radians, 4th column is timing in degree of each rectangule.

in 5th column I multiply area of each rectangule by duration of it.

example 107.92-107.51=0.409mm high of rectangle

0.409mm*117.64mm(width of retangle) * 70.02º =3371.543ºmm
0.404mm*117.64mm*69.02º=3287ºmm

ultil 0 degrees, and the sum = 83504.01ºmm

m4grity, I don't want to waste too much time understanding your spreadsheet, but I suspect you are not considering the actual open time of the transfer ports.
When you calculate the angle.area for blowdown, you are only interested in the time from EO to TO, but when calculating the transfer angle.area, you have to use the actual number of degrees from TO to TC, which is probably 130 degrees or more, not 70.

lodgernz a écrit:

m4grity a écrit:

I have this funcion on first column, I assume a retangle shape to find a maximum angle.area possible, but is to far of reality. second column is the diference of positions the high of rectangule, 3rd column is piston position on radians, 4th column is timing in degree of each rectangule.

in 5th column I multiply area of each rectangule by duration of it.

example 107.92-107.51=0.409mm high of rectangle

0.409mm*117.64mm(width of retangle) * 70.02º =3371.543ºmm
0.404mm*117.64mm*69.02º=3287ºmm

ultil 0 degrees, and the sum = 83504.01ºmm

m4grity, I don't want to waste too much time understanding your spreadsheet, but I suspect you are not considering the actual open time of the transfer ports.
When you calculate the angle.area for blowdown, you are only interested in the time from EO to TO, but when calculating the transfer angle.area, you have to use the actual number of degrees from TO to TC, which is probably 130 degrees or more, not 70.

I don't understand, when you are calculating blow down angle area, you dont care if open in 130 degree or 50 degree, you have only the reference that is blowdown timing in this case = 32º, and the transfers port will be open by ~~70 º

I'm Calcutaling 70degress = transfer open to BDC

m4grity a écrit:

I'm Calcutaling 70degress = transfer open to BDC

As I said, you must calculate transfer open to transfer close, 140 degrees

lodgernz a écrit:

m4grity a écrit:

I'm Calcutaling 70degress = transfer open to BDC

As I said, you must calculate transfer open to transfer close, 140 degrees

then assuming a rectangle transfer with 20mm , I will start in 140 degree calculate until 70 degrees (BDC) and calculate again BDC to TC (70degrees to 0) and sum? or this values are wrong?

m4grity a écrit:

lodgernz a écrit:

m4grity a écrit:

I'm Calcutaling 70degress = transfer open to BDC

As I said, you must calculate transfer open to transfer close, 140 degrees

then assuming a rectangle transfer with 20mm , I will start in 140 degree calculate until 70 degrees (BDC) and calculate again BDC to TC (70degrees to 0) and sum? or this values are wrong?

When you turn the crank 1º past Transfer Open, a certain area of port, say A1, will be exposed. I don't know or care what that area is, that's for you to measure.  That area of port A1 will be open for 140º. The angle.area for this area A1 is = A1 * 140.
Turn the crank another 1º and another area of port, A2, will be opened. Now the amount of port opened is A1 + A2.  
This area A2 will be open for 138º. The angle.area for this area is = A2 * 138.
The next degree of crank rotation will open Area A3 which will be open for 136º and the angle.area will be A3 * 136.
Keep going like this until BDC, then add up all the angle.areas you have calculated.
The total is the angle.area for that transfer port. Good luck!

ahh, thank you i have been calcutated it wrong. Thank you very much Lodgernz!

lodgernz a écrit:

When you turn the crank 1º past Transfer Open, a certain area of port, say A1, will be exposed. I don't know or care what that area is, that's for you to measure.  That area of port A1 will be open for 140º. The angle.area for this area A1 is = A1 * 140.
Turn the crank another 1º and another area of port, A2, will be opened. Now the amount of port opened is A1 + A2.  
This area A2 will be open for 138º. The angle.area for this area is = A2 * 138.
The next degree of crank rotation will open Area A3 which will be open for 136º and the angle.area will be A3 * 136.
Keep going like this until BDC, then add up all the angle.areas you have calculated.
The total is the angle.area for that transfer port.

The perfect explanation, Lodgernz .

Frits Overmars a écrit:

lodgernz a écrit:

When you turn the crank 1º past Transfer Open, a certain area of port, say A1, will be exposed. I don't know or care what that area is, that's for you to measure.  That area of port A1 will be open for 140º. The angle.area for this area A1 is = A1 * 140.
Turn the crank another 1º and another area of port, A2, will be opened. Now the amount of port opened is A1 + A2.  
This area A2 will be open for 138º. The angle.area for this area is = A2 * 138.
The next degree of crank rotation will open Area A3 which will be open for 136º and the angle.area will be A3 * 136.
Keep going like this until BDC, then add up all the angle.areas you have calculated.
The total is the angle.area for that transfer port.

The perfect explanation, Lodgernz .

Just a copy of yours Frits

Reading some old SAE papers, a question occured to me.
@Frits and Jan:
Did you ever test your cylinders with the Jante-Method?

(Comment: a couple of pivot pipes replacing the head, the engine is motored to different speeds and the pressure - equivalent the velocity of the gas is measured)

I would wonder how your results may have looked like.

Thanks!
Tim

Yes, I did, long ago, and in a very simple way!
Using a big ventilator.
I certainly learned something from it!
Later there was such a machine that worked electronically.
It was made by a Czech.
The results were not repeatable, so it was abandoned at Aprilia.

Hi Jan, Hi Frits,
Trying to understand the relation between different concepts in scavenging angles.
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As you can see 2 different cylinders but the same typ of engine.
Do you think it is normal that on the model on right side the ports 1+2 can't join together in the middle?
Here a picture from used piston with that configuration.
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Are those marked points because there is no more flow coming from 1+2 at this point, what do you think?
How to measure exactly the vertical angle of ports?
Any help would be appreciated, thank you.