alright i am doing a physics presentation tomorrow morning and i chose to do the physics of an engine. well as ive gotten deeper into it i have realized its complicated as all hell, even when trying to simplify things. this is only my second semester of physics so i dont know a bunch but enough.
the problem at the moment - im trying to figure out what is the distance between the center of the crank on a GSR to the center of the connecting rod. i have searched but couldnt find anything. the reasaon im trying to figure this out is to figure out how much rotational force is actually transfered to the crank to make it spin. right now i have figured out the force pushing down on the piston is roughly 5955lbs. and i have stated that this force happens at 10* after TDC (remember for simplicity reason, im guessing half the kids in my class only know how to fill the gas tank) so 5955 x sin(10) = 1034lbs. but im trying to figure the amount of torque applied which would be 1034 x the distance i need.

thanks

 

You mean the stroke?

 

divide the stroke by 2 right? 87.2mm/2=43.6mm

 

yeah i just started drawing skectches actually and got 4.36cm lol. thanks guys. and i think i may have the whole 10* thing wrong. when does the spark happen? before TDC or after. im guessing after bc if it was before the piston would try spinning backwards and if it was right at TDC wouldnt it push the crank out of the engine? maybe im overthinking or missing something

 

nvm again im forgetting that it takes time for the fuel to ignite.

 

it's before top dead center (BTDC), usually around 15 degrees. you want max force to occur on the piston 90 degrees AFTER top dead center (ATDC) to achieve max torque; of course you know that because you're in physics. so the fact that it takes time for combustion to occur coupled with how fast the crankshaft is spinning, 15 degrees BTDC is optimal time for spark to occur.

 

ah alright thanks alot. that was pretty much the last bit of info i needed to finish the slides.

 

90 degrees after TDC would be BDC..... why would you want max. cyl. pressure to occur then?

I've always tuned to achieve peak cyl. pressure approx. 6 degrees after TDC.
RPM and cyl. head efficiency dictate how much spark advance you need to achieve that.

 

Peak force would be delivered when the cylinder pressure is at its highest.

To determine actual POWER delivery, you'd need to integrate the cylinder pressure over a whole cycle...

Where did you come up with your 5955 lbs number?

 

Peak pressure usually occurs 3 - 10 degrees ATDC. By the time the crank is 90 degrees ATDC most of the pressure is in the cylinder is lost. Ideally peak pressure at 90 ATDC would produce the most torque but unfortunately that's not how an engine works.

Boarding2008, do you know what BMEP (Brake Mean Effective Pressure) is?

 

sc at kfi - 180 ATDC would be BDC, not 90.

shifty is right about peak power. obviously the higher the pressure, the more force is applied .... P=F/A so F=PA so as P increases so does F.

 

you're right. lol wtf was i thinking.....
180 - 360 - 540 - 720

 

I got the 5955lbs by doing 15PSIA x 10(cr of a gsr) which gives u 150PSIA. I can't remember the radius of the bore. Its like 4.05cm. Convert that to inches. Then find the area of the piston (pi r^2) times that by 150PSIA and that should give u the amount of force on the piston before the spark. The fuel ingintion increases that number (from what I have read thru research) by 3.5 to 5 times. So 5955/5 is the force pushing down before.

 

I don't know what bmep is but I'm cool with learning if your willing to teach

 

So you are assuming using the static compression ratio of the engine? I guess that's "ok" if you are assuming 100% volumetric efficiency.

 

i know this doesn't relate to the topic.. but it's nice to see real tech being talked about on here.

to the OP: are you taking calculus based physics?

 

Good. BMEP (brake mean effective pressure) is the average pressure in the cylinder over 1 cycle. It is a calculated value and is useful in determining how much power an engine can realistically make.

The formula is:

BMEP = HP x 13,000 / Displacement (in liters) x RPM

On a 4-valve engine running pump fuel the maximum BMEP you are going to see is about 185 - 200 psi (that's pretty optimistic, numbers are usually lower).

So if you had a 1.8 liter engine that makes 220 HP at 8500 rpm your BMEP would be about 186 psi.

When you are increasing the efficiency of an engine, you are increasing the BMEP. Meaning, you are increasing it's volumetric efficiency.

Sorry I can't go into more depth, I am at work and my time is limited.

Check out Wikipedia's article on BMEP I didn't read it to thoroughly but it seemed like it gives more info than I can now.

 

if it was more of an engineering course/class i would try to get as realistic as possible. but its just a basic physics course that i need. im going to school for pharmacy where physics isnt exactly needed. but its still a sick *** class lol
no just basic stuff at the moment. i have contemplated taking a more in depth physics later in a year or two. i am taking calculus and physics seperatly tho haha. and thanks for all the help guys. half the kids where dumb founded as i explained what was going on in the engine lol. presentation went well seeing that i hate public speaking.