Suzuki 2 Strokes

I am not an ingenior, I am a self made man, I have not 60 YO experience and I my english explanation is a little bit wrong ...but, I will trie to explane at every body how a motor is working.
If some of you can help me with adding pics or best explane, he is wellcome, mostly he knows better than me.
A big part of that I know is in "2 stroke tunning" about Graam Bell...just good for the beginners and in http://adardaine.free.fr/index.php?lng=en" onclick="window.open(this.href);return false;" onclick="window.open(this.href);return false;" onclick="window.open(this.href);return false;" onclick="window.open(this.href);return false; (also happyly in french....edit:not much many translation for you )...most more difficult to understand...
If this topic give more technicals questions, prehaps we will obliged to do another topic for doing not too big one...just put a link to understand a small technical question.

The crankshaft is doing a rotating movement.
The piston is doing a translate movement.
Between both, the con rod...rotating AND translate

Instead of the rpm, the piston have not a linear speed.
It does at each round
1° go from the BDC (thanks) a big acceleration.
2° before the TDC a big breaking but not so big than the acceleration (there, it takes and explosion on the face
3° stop at TDC
4° go from TDC to BDC big acceleration
5° before BDC a big breaking, biggest than the 4° acceleration
6° stop at the BDC....again immediately

We use the terms Top Dead Center (TDC) for the point where the piston is as high as it goes and Bottom Dead Center (BDC) for teh lowest point.

Piston acceleration is very low either side of TDC and BDC. It reaches a maximum a few degrees later.

tz375 wrote:Piston acceleration is very low either side of TDC and BDC. It reaches a maximum a few degrees later.

Yes but we can say the acceleration is about 1/4 of 1/2 round (from TDC to BDC by exemple) same for breaking and we can thought that the linear speed is there, between acceleration and breaking, about 2/4 of a 1/2 round.
It's also at the "linear peed" that the con rod angle is at the maximum.
So, it's around the TDC and BDC the shocks are most importants and when the speed about linear angle rod max giving also wear of the piston support.
The maximum wear we can see in a cylinder is around BDC and TDC...more ar the BDC

I do the difference between time accs and break because there is a small difference that many person does not know.

Thanks for your translation

Here, you can see the accelerate and brake and stop of the piston.
This exemple is for a 125 GT,

you can see the Speed max at 10.000 rpm...23 meters/seconds (m/s and not Ms or M/s)
But to do less difficult , I can do with theoric linear speed.
bore 43mm x 2 (for 1 round) x 10 000 rpm / 60 (for 1 second) /1000 (for 1 meter) = 14 m/s

<--------------biting her tongue..............ARRRGGHH!!!!



Suzukidave.......Love your caricature!!!!!

There are really several issues here. The usual measurement of piston speed is Mean Piston Speed which in the case of say a T500 @10.000 rpm would be:

2 * (64 / 1000) * 10,000 / 60 = 21.3 m/s. It is generally accepted that somewhere around 25m/s (4900ft/min) is about as fast as a piston can travel, though some F1 engines have approached 28m/s (5500 ft/min). Back when the GT was designed, anything over 3500 ft/min (17.8m/s) was considered really high and that works back to about 8,400 rpm.

Modern pistons and rods could probably exceed those levels for a short time.

That's a great way to understand one aspect of the motor but does not tell us anything about peak velocity or piston acceleration. The actual velocity does follow that sinusoidal curve already posted and it's easy to see that teh velocity and acceleration are zero at the points of inflection (TDC and BDC) The piston accelerates away from those points quite slowly at first as can be seen from the graph.

That's all relevant to piston movement. If you wish to calculate loads on the big end or in the con rod there are algorithms for that too and the load is proportionate to mass and to the square of rpm. Double the piston mas and the load doubles. Double engine speed and load increases by 4 times (2^2).

so whats my 04 honda 600rr piston doing at 14,000rmp. haha

crash277 wrote:so whats my 04 honda 600rr piston doing at 14,000rmp. haha

Busting a move .. not as detailed as Teasers comment , but its as close as i can get

crash277 wrote:so whats my 04 honda 600rr piston doing at 14,000rmp. haha

Exellent exemple (but this piston is VERY short) but your bore longer (don't know the name in english) is 42.5mm , about the same than the 125 GT and it does 20 m/s at 14000rpm.

I don't know about you , but as I am not an engineer,
_m/s don't speak to me...I prefer km/h or for you mph. So, to have an idea 20m/s linear speed are 72km/h or 45 miles per hour....all that into 42,5mm longer with breakings and accelates.
_ rpm don't speak to me, I prefer round per second 14000 rpm are 233 rp/s
So , it does in one second, 233 rounds and 466 breakings, and 466 accelerates and all that at 45 mph average...and in this one second, this piston deplacement is 42.5mm/1000 x 2 x 233 = 20 metres longer (m/s)
....whith those numbers I undertand better what's happen in a motor (here 600rr).

Crash, as Dino said, it's 19.8m/s or just under 4000 ft/min. That's the same as a GT750 running at 9,300 RPM

That's a classic example of how to get more power. Go to more cylinders with shorter stroke and let it rev.