I'm Have been looking for this for a while and thought I might ask here.
Does any one have a link to a site with this info?
If any one can give me a formula for this you will be the smartest person I know.

I will search some more and if I find it I will post it here and on my cardomain site .

 

Remember F=Ma?

That's the foundation for finding what you're after. You need to calculate piston acceleration. You can simplify things a bit by working the formula at TDC, where acceleration is highest.

From there you just need to convert units and multiply that by your reciprocating mass to find your "load".

F=Ma

M = reciprocating mass
a = piston acceleration (g's)

Well.... those are the basics of it.

edit: And keep in mind that those forces are not on the "piston"... but the pin and rod bolts (the joints of the assembly).

-Mindgame

 

Just to add to what mindgame said.

Remember that mass is not the same as weight. In the english system, I weigh 165 lbs. I have a mass of 165/32.2 = 5.12 slugs. 32.2 is the acceleration of gravity in ft/s^2. If you want to know how many g's, divide out the acceleration of gravity after you find the acceleration of the piston at TCD. For example, if you find that the acceleration of the piston is 64.4 ft/s^2, then that would be 64.4/32.2 = 2g's.

F=ma!

 

"Piston 4.180" around 400g
Pin, shorter and smaller around than street stuff, around 75g
Locks, 3g
Rods, They are limited down to 525g, so the recipricating mass there is light let's say 150g
Rings, narrow rings don't just cause less drag but have less weight too. So around 35g for rings.

663g vs the 780g on the normal stuff.

The piston speed of a Winston Cup motor at 9500 is about 5162ft/min, the piston g's are 5260.

.663g x 5260 x 9.81 = 34211 N = 7690 lbs of F

So that's a good amount of Force working to pull the piston out of the rod."

That's a little help on what goes on. Piston G's are harder to get, math I let computers do for me rather than try and figure it out.

Bret

 

I'm was looking for the to find out how much weight the rod holds back. At the pin/rod sm. end and how much the bolts hold back.
I'll have the weights of all the parts because I will have them on hand soon.
I searched last night and found a lot of weird stuff, but no piston g's.

 

The easiest way, as SStrokerAce said, is to let an already written program do the calcs. If you have engine stroke, rod length and max rpm I'll give you max piston gs. You could then estimate loads using F=ma.

The rod weight (mass) less the cap, significantly adds to the load on the rod bolts, but not to the pin.

FWIW, why do you want to know the exact loads?

Remember that the loads are directly proportional to weight (mass) of the parts and proportional to the square of the rpm. IOW, doubling the mass of the parts doubles the gs and therefore the load, but doubling the rpm increases the gs and load by 4 times. Thats why 19000 rpm F1 pistons see 10000 gs.

 

A 383 SBC with 6" rod and a normal weight 4.030 piston.
Speed 7000-4375 ft/s
Speed 7500-4688 ft/s

G's 7000-3420
G's 7500-3930

All well within the limits of good part's.

 

How about a 3.4L engine.
3.31'' stroke
5.7'' rod
Now a Have a paper on how to find piston in bore hight and rod to bore angle. Would that be part of it?
How about those parts at 6500, 7000, 7500 and 8000?

 

Calculated piston gs for 3.31 stroke and 5.7 rod:

6500 = 2560 g
7000 = 2970 g
7500 = 3410 g
8000 = 3880 g

Compare these to the numbers 1racerdude got for 3.75 stroke and 6.0 rod.

 

Your freakin awsome
How do you do that?

 

I cheat. It's not what you know, but where to find it that counts.

Try these links to get an idea of the simulation program. Piston Gs @ TDC is an output.

http://www.performancetrends.com/Eng...r_Pro_v3.3.htm

http://www.performancetrends.com/PDFs/EAPSupp4.pdf

 

That's not cheating,thats just having the right tools for the job.
Beats the hell out of a ledger sheet and a pen.

 

How do you calculate g force? thats easy hang a pieceof string from your rearview mirror attach a lead sinker to the end when the string swing's 90* from vertical your at 1G

 

Are you sure about that?

Draw a moment diagram with the weight of the sinker pulling it down toward the ground (or center of the earth), aka gravity, and a force equal to it's weight (the 1.0 lateral g) pulling it 90 degrees from down. Now isn't the resultant force on the sinker at 45 degrees from vertical and 1.414 times the weight of the sinker? Won't the sinker be at 45*, not 90* @ 1 lateral g?

PS: I tried the sinker thing on the top of a piston to check gs but experienced a few problems even before I got to 6500 rpm. I now have a Pb coated piston.

 

How much is that soft ware?
It didn't show a price.