[GP125] All that you wanted to know on Aprilia RSA 125, and more, by Mr Jan Thiel and Mr Frits Overmars (PART 5)

Frits Overmars a écrit:

Your cylinder produces 7-9 hp at 11000 rpm with a blowdown time.area that is only 1/3 of what it should be for 11000 rpm (11000 / 3560 = 3,09), and the same is true for the power it produces at 11000 rpm; it makes perfect sense.

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Ciao Frits, I miss something. You say that angle area is 1/3 of what it should be (clear) and so also max Power is 1/3 of what it should be (clear) but how do you get that this engine with optimum angle area would give 21/29 HP @ 11000?

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I wrote something wrong: angle area is 1/3 of Ideal, but power is 1/3 of what it should be because of ratio rpm-real-max-power/rpm-AA-max-power (11000/3600)

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JanBros a écrit:

Frits, in this drawing Blowdown is continuing a bit while the transfers are already open (as it is in real life), but I wonder if this has to be incorporated into calculations or not ? How much would you say on average (you are probably going to say "can not tell you, it depends on the preasure still left in the cylinder at TO).
taking measures from the drawing, it's about 2.6°. I can manage to re-create the RSA in my Excell meeting the STA for transfer with the data from the RSA drawings , but with loosing those 2.6° of the transfers, it would be impossible.

Your 2,6° estimate is spot-on, Jan.
The geometrical exhaust timing is 101° before BDC till 101°  after BDC
The effective  blowdown timing is  95° before BDC till   63° before BDC

The geometrical transfer timing is 66° before BDC till 66° after BDC
The effective    transfer timing is  62° before BDC till 64° after BDC

The geometrical values are fixed; the effective values change with rpm. The higher the rpm, the more will the effective values deviate from the geometrical values. These deviations are established in the simulation software.
In other words: I do not tell the sim what they should be; the sim tells me. Oh, the joy of a well-mannered sim .

Citation :

You also said "our engines should meet the same STA's as the RSA" . Does this count for ALL/any engine or only for thermaly healthy high BMEP engines ?

Let's begin with saying that all engines should be as thermally healthy as the RSA. Only then can we think about meeting the other RSA qualities. Having said this, the specific time.area values are supposed to be universal.

Citation :

I would think you'd say yes (on thermaly healthy high BMEP) , but am in doubt as less rev's gives higher STA., but less BMEP would require less A.A so levelling things a bit.

Lower revs give a longer port open-time during each revolution, so the actual time.area increases, although the required time.area should remain constant, which means, as you say, that the required angle.area can be smaller for lower revs.

Citation :

Can one have to much transfer T.A.?

Yes you can, for two reasons.
During the transfer phase the mixture flow through the transfer ducts accelerates as long as the crankcase pressure is higher than the cylinder pressure. After that, this flow is slowed down because the crankcase pressure has now dropped below the cylinder pressure.
As soon as the transfer flow comes to a halt, the transfer ports should close; otherwise the pressure differential, helped by the rising piston, will reverse the mixture flow.
The second reason that you can have too much transfer time.area is that you need sufficient blowdown time.area above the transfer ports. If the transfer ports are higher than they need to be, the exhaust port must also be higher than it would otherwise need to be.

Tnx Frits.

so basicly : a standard Sky cylinder has to high transfer timings to produce max power at 8.000 rpm and I should lower them (and make them wider) Try explaining that to teenagers on their scooters

flatart a écrit:

I wrote something wrong: angle area is 1/3 of Ideal, but power is 1/3 of what it should be because of ratio rpm-real-max-power/rpm-AA-max-power (11000/3600)

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I think he meant, that if you had that STA matching RSA levels at 11,000rpm, you could assume that 29HP. Aprilia has about 59HP at crank @13,000, and it's 125cc. So that is about 36,3HP/1000cc/1000rpm. If one has 74cc, it should produce about 0,074*11*36,3=29,5HP @11,000rpm, IF it is as good and effective as RSA.

op76 a écrit:

I think he meant, that if you had that STA matching RSA levels at 11,000rpm, you could assume that 29HP. Aprilia has about 59HP at crank @13,000, and it's 125cc. So that is about 36,3HP/1000cc/1000rpm. If one has 74cc, it should produce about 0,074*11*36,3=29,5HP @11,000rpm, IF it is as good and effective as RSA.

thanks for your clear explanation, but what's STA?????

Specific Time.Area it transforms the T.A of your engine into a universaly comparable number between different engines like BMEP.

it is T.A divided by the volume above BDC ( so cyl. + combustion chamber) multiplied by 1000; resulting in HP/1000cc/1000rpm

Ciao Frits, I have a very difficult question, for wich I think there is no exact answer. I know that you have no magic stick but at least you or someone else can tell me how far from reality I'm with my findings. I'm trying to write down the RSA pressure curve inside the cylinder, from the spark to ports opening.
I know that it depends on a lot of factors but I'm interested just on the basic shape and some reference values. Some assumptions: full gas and max torque rpm.
So I read somewhere that the peak-pressure should be arond 15/20° after TDC; I read somewhere else that this peak for RSA should be 65bar and I read somewhere else that at exhaust opening it's about 12bar.
As far as the shape curve is concerned, I took it from a 2-stroke book.
So this is my RSA-pressure curve in blue; in orange you can see the "resulting" pressure considering all the angles between piston, conrod, crankshaft: I mean the force with which the piston pushes down is proportional to the cylinder pressure, but this force when it's transmitted to the crankshaft has 2 components: one component is "radial" is trying to destroy crankshaft bearings (ironic); the other component is "tangential" and it's the one that we love because it produces torque.
So the orange curve is considering the tangential component, I know that I'm mixing Pressure with force but they are proportional, so it's ok.
I have calculated torque produced with orange curve doing some simplification, just considering 80° of pushing phase (exhaust open at 80 ATDC) and not considering the mechanical loss between 80 and 360° and I got about 30Nm that it's quite similar to 32Nm of max torque of the RSA, so it means I'm not so far..
Why I'm doing this exercize? I've tome to waste and I just want to better understand what's going on inside the combustion chamber and how this pressure is converteded in torque considering engine geometry.



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flatart a écrit:

Ciao Frits, I have a very difficult question, for wich I think there is no exact answer.

Flavio, the question is not difficult; the problem is posting the answer. I see no option for attaching text files, so the only solution seems to be dumping the pressure list right here. It shows the absolute cylinder pressure from BDC to BDC.
The pressure history during open exhaust port is distorted, but the pressure data during closed exhaust port should be fairly accurate. The maximum combustion pressure is 124,7 bar @ 16° after TDC.

.1553192
.1819933
.2086675
.2353416
.2620157
.2886899
.315364
.3420381
.3687122
.3953864
.4220605
.4487346
.4754088
.5020829
.528757
.5554311
.5821053
.6087794
.6354535
.6621277
.6888018
.7154759
.7421501
.7688242
.7954983
.8221725
.8488466
.8755207
.9021949
.9288689
.9555431
.9822173
1.008891
1.035565
1.06224
1.088914
1.115588
1.142262
1.168936
1.19561
1.222284
1.248959
1.275633
1.302307
1.328981
1.355655
1.382329
1.409003
1.435677
1.462352
1.489026
1.5157
1.542374
1.569048
1.595722
1.622396
1.649071
1.675745
1.702419
1.729093
1.755767
1.782441
1.809115
1.835789
1.862464
1.889138
1.915812
1.942486
1.96916
1.995834
2.022508
2.049182
2.075857
2.102531
2.129205
2.155879
2.182553
2.209227
2.235901
2.262575
2.28925
2.315924
2.342598
2.369272
2.395946
2.42262
2.449294
2.475969
2.502643
2.529317
2.555991
2.582665
2.609339
2.636013
2.662688
2.689362
2.749123
2.811325
2.876093
2.943555
3.013852
3.08713
3.163543
3.243259
3.326453
3.413313
3.504038
3.598842
3.69795
3.801606
3.910065
4.023604
4.142514
4.267112
4.397732
4.534731
4.678495
4.829436
4.98799
5.15463
5.329859
5.514223
5.708302
5.912718
6.128144
6.355296
6.594944
6.847925
7.115128
7.397508
7.696104
8.012017
8.34645
8.70067
9.076062
9.474103
9.896379
10.3446
10.82058
11.32629
11.86381
12.43538
13.04339
13.69036
14.37898
15.1121
15.8927
16.72392
17.60903
18.55141
19.55453
20.62192
21.75709
22.96352
24.24457
25.60333
27.04256
28.56457
30.17094
31.86242
33.63871
35.49807
37.43721
39.45081
41.53142
43.66889
45.8503
48.05958
50.27749
52.4814
54.64566
56.74171
58.73851
60.60363
62.80111
65.32521
68.16091
71.28336
74.65852
78.24346
81.98818
85.83744
89.73278
93.61511
97.42722
101.1155
104.6322
107.9366
110.9962
113.7866
116.2915
118.5025
120.4181
122.0427
123.3857
124.4604
124.7206
124.2303
123.0984
121.4243
119.2984
116.8008
114.0031
110.9675
107.7483
104.3924
100.9395
97.42342
93.87267
90.31075
86.75714
83.22768
79.735
76.28912
72.89774
69.56656
66.2997
63.09985
60.07553
57.23529
54.56703
52.05932
49.70148
47.4836
45.39634
43.43103
41.57956
39.83441
38.18856
36.63548
35.16916
33.78391
32.47453
31.23613
30.0642
28.95453
27.9032
26.90659
25.96131
25.06421
24.21236
23.40304
22.6337
21.90197
21.20564
20.54266
19.91109
19.30915
18.73515
18.18754
17.66483
17.16566
16.68875
16.23289
15.79696
15.3799
14.98072
14.59848
14.23233
13.88143
13.545
13.22234
12.91276
12.6156
12.33028
12.05621
11.40806
10.86002
10.37817
9.944427
9.547805
9.181016
8.838936
8.517795
8.214724
7.927478
7.654261
7.393608
7.144308
6.905341
6.675845
6.455076
6.242394
6.037238
5.839115
5.647591
5.462277
5.282827
5.108931
4.940306
4.776698
4.617875
4.463627
4.313758
4.168092
4.026465
3.888725
3.754734
3.624362
3.497487
3.373998
3.25379
3.136765
3.022833
2.911907
2.803907
2.698758
2.596388
2.49673
2.399722
2.305304
2.213418
2.124013
2.037039
1.952446
1.87019
1.790228
1.71252
1.637027
1.563712
1.49254
1.423479
1.356496
1.291562
1.228649
1.167729
1.108777
1.051767
.9966784
.9434862
.89217
.8427114
.7950888
.7492867
.705286
.6630707
.6226263
.5839376
.5469904
.5117712
.4782696
.4464712
.4163675
.3879471
.3612003
.3361177
.3126917
.2909145
.2707787
.2522774
.235405
.2201557
.2065248
.1945085
.1841011
.1753006
.1681042
.1625099
.158514
.1561184
.1553192


Fantastico, grazie Frits.
How ho you get such a detailed curve??!!

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Flavio, here is some more documentation regarding gas pressure, inertial force, resultant force. and torque.
Forces above the horizontal center line are pushing forces in the con rod; forces below this line are pulling forces.
The blue horizontal line represents average torque during one crankshaft revolution.
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Frits these last infos Will take a lot of my time in the next days. Dont know how to thank you for sharing such info, it's like chatting with Einstein about relativiy, but 2 stroke is quite more funny, by far

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flatart a écrit:

Frits these last infos Will take a lot of my time in the next days. Dont know how to thank you for sharing such info, it's like chatting with Einstein about relativiy, but 2 stroke is quite more funny, by far.

My pleasure, Flavio

Frits Overmars a écrit:

Flavio, here is some more documentation regarding gas pressure, inertial force, resultant force. and torque.
Forces above the horizontal center line are pushing forces in the con rod; forces below this line are pulling forces.
The blue horizontal line represents average torque during one crankshaft revolution.

Sorry can you explain what the inertial force in the graph stands for?

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flatart a écrit:

can you explain what the inertial force in the graph stands for?

It's the force that would be necessary to accelerate the piston from zero velocity in the TDC and BDC positions to its maximum speed, and to slow the piston down again when it is nearing the next BDC and TDC positions, when there is no compression or combustion pressure acting on the piston.
Without this pressure, the con rod must pull the piston down from TDC until it reaches its maximum speed, then push against the piston until comes to a stop in BDC, keep on pushing until the piston reaches its maximum speed in the opposite direction on the next up-stroke, and finally pull at the piston until it comes to a stop in TDC again.

This necessary force is dependent on rpm and on the total reciprocating mass of piston, ring, gudgeon pin, small end bearing and con rod.
The table below shows the influence of different reciprocating masses.
It compares the normal situation (white) with a situation where the reciprocating mass is doubled (red).

The ratio of crankshaft stroke to con rod length also has an influence on the inertial forces.
Below is a comparison of the normal 120 mm con rod with a 60 mm con rod (in reality a 60 mm con rod could never be used because in BDC the piston would hit the crankshaft).

Frits Overmars a écrit:

Flavio, here is some more documentation regarding gas pressure, inertial force, resultant force. and torque.
Forces above the horizontal center line are pushing forces in the con rod; forces below this line are pulling forces.
The blue horizontal line represents average torque during one crankshaft revolution.

these graphs really enclose so many infos that it will take days or weeks to get everything is hidden in them.
The simplest one is that the greatest the inertial force (mass of crankshaft, flyweel and everything connected), the greatest will be the opposition to acceleration in the "pushing" phase (from TDC to EXhaust opening, 79° ATDC), but in the "not-pushing" phase (from 79° to TDC) the greatest will be the opposition to friction loss and also to pressure rising between exhaust closing and TDC (from 281 to TDC).
So first problem: find the right compromise for crankshaft weight, flyweel weight etc... Obviously this can be done with dyno-test, I guess if there's a way to do some kind of calculations (I think everything is hidden in these graphs).

curiosity:
- considering an engine revving at a fixed RPM, how does the angular speed vary in one revolution of the crankshaft? I guess it accelerates between TDC and exhaust opening and decelerates in the remaining part of the revolution. But how much in %? just to have an idea...

- have you ever tried a compression test on RSA engine with pressure gauge plugged in the spark-plug? I mean test usually done on old car egnines to check if compression is still good. What's the peak value you get on RSA cylinder?

However after trying to go deeper in these topics I really do understand Thiel's empirical approach.

flatart a écrit:

considering an engine revving at a fixed RPM, how does the angular speed vary in one revolution of the crankshaft? I guess it accelerates between TDC and exhaust opening and decelerates in the remaining part of the revolution. But how much in %? just to have an idea...

The pictures below may help. They show the angular speed fluctuation and the crank angle fluctuation within one crankshaft revolution for 6500 rpm and for 13000 rpm (I used the same pressure history for the 13000 rpm situation and the 6500 rpm situation in order to exclude its influence. In reality the pressures at 6500 rpm would of course be much lower).

You will notice that at 13000 rpm the fluctuations are much smaller than at 6500 rpm.
The energy needed to accelerate the rotating masses within one revolution rises with the square of the rpm and this energy must be administered in a shorter period of time as the rpm rises.
In other words: a high-revving flywheel can be much smaller than a low-revving flywheel and still bring about the same smoothing effect.

Citation :

have you ever tried a compression test on RSA engine with pressure gauge plugged in the spark-plug? I mean test usually done on old car engines to check if compression is still good. What's the peak value you get on RSA cylinder?

No, we never did that. But I do use the contraption shown below for stationary leak tests, in order to see if the piston ring is still moving freely or pinched in its groove.
Pressure is applied via the spark plug hole. Then you rotate the crankshaft until the exhaust port closes.
If the piston ring is pinched, you can even establish on which side of the piston the pinching has happened, because the slanting con rod will push the piston against one side of the cylinder bore. If that is the side opposite the pinching, the piston ring will be lifted off the bore, causing a pressure leak.

In case you are still curious about the result of a car-type compression test, we can perform a simple calculation.
These tests are usually carried out with no more than 200 rpm, which means that the effective exhaust timing is about the same as the geometrical exhaust timing: from 101° before BDC till 101° after BDC. With a 54,5 mm stroke and a 120 mm con rod length this means an exhaust port height of 29,4 mm.
With a 54 mm bore, the cylinder volume above the exhaust port is 54*54*PI/4 *(54,5-29,4) = 57,485 mm³ = 57,5 cc.
The combustion volume is 8,6 cc, so the compression ratio above the exhaust port is (57,5 + 8,6) / 8,6 = 7,7.
The pressure in the cylinder is 1 bar absolute. According to the gas law for adiabatic compression, which states that pressure * volume^1,4 is constant, we find that the compression pressure will be 17,4 bar.

Good evening Frits,
Thanks again for your high level contributions. I'm sorry I have a certainly lower level question: I'm actually building a vintage MX bike 50cc based on a Minarelli P6 engine long stroke (38,8 x 42mm).
I have a Gilardoni cylinder with 180° exhaust timing and 130° transfer, which does not look so bad. But this is only a 3 transfers cylinder and when I calculate the angle.areas and the subsequent max torque rpm, I find only about 9500 rpm (fortunately both blowdown and transfer are aimed to the same rpm).
The intake equivalent diameter is 26mm, and exhaust is about the same. I have a 28mm carb, and I need to build the pipe from scratch.
I know aircooled engines with piston-port itake are not your cup of tea, but what would you recommend as the best guideline direction to build my pipe? Should I try to match the 9500 rpm for max torque, or should I try a compromise with a slightly shorter pipe that would also match with the 28mm carb?

carlovitch1 a écrit:

Good evening Frits,
Thanks again for your high level contributions. I'm sorry I have a certainly lower level question: I'm actually building a vintage MX bike 50cc based on a Minarelli P6 engine long stroke (38,8 x 42mm).
I have a Gilardoni cylinder with 180° exhaust timing and 130° transfer, which does not look so bad. But this is only a 3 transfers cylinder and when I calculate the angle.areas and the subsequent max torque rpm, I find only about 9500 rpm (fortunately both blowdown and transfer are aimed to the same rpm).
The intake equivalent diameter is 26mm, and exhaust is about the same. I have a 28mm carb, and I need to build the pipe from scratch.
I know aircooled engines with piston-port itake are not your cup of tea, but what would you recommend as the best guideline direction to build my pipe? Should I try to match the 9500 rpm for max torque, or should I try a compromise with a slightly shorter pipe that would also match with the 28mm carb?  

Carlo, I assume the inlet on your vintage engine is piston-controlled. If you can keep the inlet timing down to no more than 160°, a 28 mm carb should be able to function. You might also try varying the inlet length between carb and cylinder in order to adapt the power characteristics for MX use.
Forcing the engine to run at a higher rpm than the angle.areas permit, by means of a shorter pipe, will probably give less power, reduce its reliability and cause an engine character that is unsuitable for MX.

Thanks Frits for your quick answer ! I have 162° inlet timing, but this is due to a cut I increased at the bottom of the piston skirt, I will use another piston and cut less, and also build my pipe for the 9500 rpm.
Thanks again !

Frits Overmars a écrit:

In case you are still curious about the result of a car-type compression test, we can perform a simple calculation.
These tests are usually carried out with no more than 200 rpm, which means that the effective exhaust timing is about the same as the geometrical exhaust timing: from 101° before BDC till 101° after BDC. With a 54,5 mm stroke and a 120 mm con rod length this means an exhaust port height of 29,4 mm.
With a 54 mm bore, the cylinder volume above the exhaust port is 54*54*PI/4 *(54,5-29,4) = 57,485 mm³ = 57,5 cc.
The combustion volume is 8,6 cc, so the compression ratio above the exhaust port is (57,5 + 8,6) / 8,6 = 7,7.
The pressure in the cylinder is 1 bar absolute. According to the gas law for adiabatic compression, which states that pressure * volume^1,4 is constant, we find that the compression pressure will be 17,4 bar.

I did the calculation for my engine and I get 13,5bar and doing compression test with pressure gauge I have 12 bar. I guess the difference is due to the fact that piston and piston Rings are not perfectly hermetic.
So even if I had the same perfect ducts efficiency or RSA (just a dream...) pressure inside my cylinder would be 77,5% of RSA cylinder pressure (13.5/17.4). I think I can say with great aproximation that pressure is about 50% of RSA cylinder.
Grazie Frits

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carlovitch1 a écrit:

Thanks Frits for your quick answer! I have 162° inlet timing, but this is due to a cut I increased at the bottom of the piston skirt, I will use another piston and cut less, and also build my pipe for the 9500 rpm. Thanks again!

162° inlet timing might still work, so you could try with the present piston first. And when you fit the new piston,
you could try shortening it in small steps, for example 3° at a time. That way you will learn a lot.

flatart a écrit:

Frits Overmars a écrit:

According to the gas law for adiabatic compression, which states that pressure * volume^1,4 is constant, we find that the compression pressure will be 17,4 bar.

I did the calculation for my engine and I get 13,5bar and doing compression test with pressure gauge I have 12 bar.
I guess the difference is due to the fact that piston and piston Rings are not perfectly hermetic.
So even if I had the same perfect ducts efficiency or RSA (just a dream...) pressure inside my cylinder would be 77,5% of RSA cylinder pressure (13.5/17.4).

Piston and piston rings are never 100% hermetic, as you say; this is one reason for the difference between the measured pressure and the calculated pressure.
There is also a second reason. Above, I used the formula for adiabatic compression (pressure*volume^1,4 is constant)
which assumes that there is no heat being transferred into or out of the gas during the volume change.
But the compression raises the gas temperature, and some of this heat will be transferred to the surrounding metal.
You can compensate for this heat loss by using a lower coefficient, for example 'pressure*volume^1,35 is constant'.
With your compression ratio of 6,42 above the exhaust port that would give a compression end pressure of 12,3 bar.

The shape of the transfer ducts has no influence whatsoever on the static compression measurement of a two-stroke engine because the in-cylinder pressure is 1 bar absolute, until the exhaust port closes and the compression begins.

[quote="Frits Overmars"]

carlovitch1 a écrit:

Thanks Frits for your quick answer! I have 162° inlet timing, but this is due to a cut I increased at the bottom of the piston skirt, I will use another piston and cut less, and also build my pipe for the 9500 rpm. Thanks again!

162° inlet timing might still work, so you could try with the present piston first. And when you fit the new piston,
you could try shortening it in small steps, for example 3° at a time. That way you will learn a lot.

OK, thanks again Frits, I will do so.
I have another cylinder that may fit, and it has 188° exhaust timing and no significant increase in transfer angle.area (I measured it at 131° instead of 130 for the exact same area). Thus the blowdown angle.area is aimed to a significant higher rpm than the transfer (somewhat 12000 rpm versus 9700). The point is that I cannot increase the transfer area or very little otherwise I will have the piston ring locating pin going inside of it.
Do you think it is usable?

Frits Overmars a écrit:

The shape of the transfer ducts has no influence whatsoever on the static compression measurement of a two-stroke engine because the in-cylinder pressure is 1 bar absolute, until the exhaust port closes and the compression begins.

Yes I know, I didnt mean cylinder static pressure but pressure during combustion.
This depends on static compression and on the quantity of fresh charge available for combustion. The latter depends on ducts efficiency to fill the cylinder.
So if RSA peak pressure is 124bar, in my cylinder might be around 60bar (great aproximation) while I can assume that the shape of pressure curve is the same (again great aproximation)

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