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Madd Tyte JDM yo ®
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Discussion Starter #1
ok, im just trying to do some rough calculations to figure out what kind of forces are applied to the rod bolts during peak RPM usage. i took some weights and speed measurements and did some conversions. now, i didnt get the weights of the rings so i wasnt able to incorporate that or the rod bearings. i just want to see if anyone can check my math to see if i did everything ok.

Bore: 83mm
Stroke: 91mm
Piston speed: 4776 (ft/min)@ 8000rpm or 24.26(meters/sec) @ 8000rpm
Rod length: 151.892mm or 5.98in

Rod weight (2jz): 757 grams
Piston weight (2jz): 377 grams
Piston pin weight (2jz): 133 grams
Piston rod combo wt (2jz): 1267 grams or 2.793 lbs
Rod bolt diameter: 10mm x 2

Fc = mv2/r, where Fc = centrifugal force, m = mass, v = speed, and r = radius.

To perform the calculations we have estimated that the piston rod combo has mass of 1.267 kg, the radius of the circle (Crank throw) is 45.5 mm, which is 0.045 meters, and the rotational speed is 38.11 meters per second 8000rpm. each crank revolution covers Pi*D, or 3.14*91mm=285.88mm per Rev. 0.28588 meters * 8000 revolutions=2287 meters. if you can cover 2287 meters in one minute, then you cover 38.11 meters in one Second.

Step 2: The centrifugal force is calculated (see calculation below) to be 2146.016 Newtons.

Centrifugal Force = [Mass x (Velocity*2)]/Radius=Force in Newtons

Fc = {1.267 (kg) * {38.11 (m/s)}2}/ 0.045(m) = 2146.016 (N)

This force of 2146.016 Newtons is applied to two rod bolts of 10mm diameters each. the Cross sectional surface area of a 10mm bolt is 78.5sq.mm, or 157sq.mm total. Force applied = 2146.016N/157sq.mm to get this into standard units, or zeroed out to Lowest common denominator, i would divide 2146.016 by 157sq.mm and come out with 13.668 Newtons per square millimeter. If i convert 13.668 N/mm2 to PSIa, it comes out to be 1982.375.
so, at 8000rpm, a piston rod combo weighing close to the 2jz setup applies a force of 1982.375 psia to the rod bolts. does this sound ok?

thanks
chris
 

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I heart 80's Toyota's
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the accuracy of the answer is going to depend greatly on your estimation mass and crank throw... I'll guess that tossed the piston and rod on a scale... but what did you do to measure the radius of the crank?
 

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Madd Tyte JDM yo ®
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Discussion Starter #3
actually, the weights were from Dusty at MVP cuz he weighed the stock 2jz components to compare them against the HKS 3.4L stroker kit. the stock 2jz parts are heavier than the HKS parts, coincidentally.

the stroke of the motor is 91mm, and the throw of the crank is half that. the throw of the crank at the top and then to the bottom is equivalent to the 91mm stroke from TDC to BDC.
 

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This will be the amount of force if you are runinng the engine w/o compression resistane. It will change when you add that.
 

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Remember, you are calculating centripetal force. This is the force that always points towards the center of the circle of rotaton. (i.e, inwards)

There are only two instances in which the full magnitude of the centripetal force is applied to the rod bolts, they occur at TDC and BDC.

If you want a more accurate representaiton of the forces applied to the rod bolts, you need to take into account the forces that occur when the mass of the piston and rod change direction at TDC and BDC.
 

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Madd Tyte JDM yo ®
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Discussion Starter #6
well, the compressive loads during compression and the power stroke are not really relavant to the tensile strength of the rod bolts. the tensile strength is more related to the change of direction of the piston in the bore as the piston starts to slow down as it nears TDC during the exhaust stroke and beginning of the intake stroke.

TDC to BDC intake stroke= tensile
BDC to TDC compressoin stroke = compressive
TDC to BDC power stroke = compressive
BDC to TDC exhaust stroke= compressive till mid-stroke as the piston speed slows down.

this is tough and my brain is tired now... lol
 

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Madd Tyte JDM yo ®
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Discussion Starter #7
the formula was for CentRIFugal force, not centripital. i was calculating the force pulling outward from the center. i made sure not to calculate the wrong thing.

i also was under the impression that the biggest load was when the piston reaches just past the midway point in the cylinder. it experiences maxiumum accleratoin up to mid-bore and has reached maxiumu speed. as soon as it passes the mid point, it start to decelerate from max speed and i read somewhere that this was the point of peak tensile load, assuming no compressive forces applied upon the piston such as compressoin stroke or power stroke.
 

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flubyux2 said:
the formula was for CentRIFugal force, not centripital. i was calculating the force pulling outward from the center. i made sure not to calculate the wrong thing.
No my friend... mv^2/r is the equation for centripetal force.

Centrifugal force is simply F=ma, where a is calculated from centripetal force.

Look it up...
 

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Madd Tyte JDM yo ®
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Discussion Starter #9
i did look it up. thats why i said it WAS for centrifugal. i did a search for centrifugal force calculation andtahts what it gave me. oh well... guess i cant count on search engines to lead me to right information. dammit.
 

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Nautical Blue MR2T
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A post of mine that I copied directly from the MR2oc.

8500 RPM = 141.6 Revolutions Per Second.
1 Second / 141.6 RPS = .007 (Time for one revolution).
.007 / 2 = .0035 (Time it takes for one stroke).
.0035 / 2 = .00175 (Time it takes to get from TDC to 90 degrees).

So: ( /\ is supposed to be that triangle symbol that means "Change in").

T = .00175s
/\X = .043m (Half the stroke).
Vo = 0 m/s

(Kinematic Equation):
/\X = VoT + (1/2)A(T^2)

.043 = 0 + (1/2)A(.0000030625)

(2 x .043) / .0000030625 = A = 28081.6 m/s^2. (meters per second - per second).

NOW onto the actual forces. (F = ma. Force in Newtons = Mass times Acceleration).
CP / Eagle combo mass: 1056g = 1.056 kg.
OEM combo mass: 1299g = 1.299 kg.

For the CP / eagle combo:
N = 1.056kg x 28081.6 m/s^2 = 29654.17N

For the OEM combo:
N = 1.299kb x 28081.6 m/s^2 = 36647.00N

29654.17N / 36647.00N = .8.


Thats a 20% decrease in tensional forces on the rod bearings and rod caps by going with Eagle rods vs OEM rods. However. This number would increase since with the whole rod / stroke ratio thing making the piston have to move further going from TDC to 90 degrees than ideal, it would make the peak acceleration greater, and thus the forces would be greater and the difference would be more pronounced.

Conclusion
Eagle Rods: 560g
OEM Rods: 773g

CP pistons: 398g
OEM pistons: 398g

CP wrist pin: 98g
OEM wrist pin: 128g

Difference in tensional stress (the force that tries to pull the rod off the crank): 20%

Am I glad I spend the little bit of extra money (less than 100 bucks) on the Eagles vs OEM rods with ARPs? YES

I mentioned that the piston accelerates at about 28081.6 m/s^2. For reference, Gravity accelerates things at 9.8 m/s^2.... People can die if accelerated more than about 98 m/s^2. The piston is still taking about 287 TIMES that amount of acceleration. I think it should be a given to try to keep the forces associated with that acceleration at a minimum.
 

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Madd Tyte JDM yo ®
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Discussion Starter #11
awesome, thats great info. looks like i pulled up the wrong formula. i dont know why it gave me the centripital force even though i searched centrifugal. it makes sense tho cuz its straight outward force.
 

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Nautical Blue MR2T
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Technically, the force on the rod bolts at the start of the intake stroke would actually be more due to the piston having to accelerate faster toward peak piston speed due to the rod / stroke effect of the crank trying to pull the piston to the side, but the piston can only go down, so it has to go down faster. But yeah, for the most part, without the factor that the crank is rotating (very minor considering the forces already working), those numbers show the advantage of lighter engine components.

The thread was entitled something like "The rear advantage of Eagle Rods!" or something like that on www.mr2oc.com .
 

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Madd Tyte JDM yo ®
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Discussion Starter #13
yeah, i can see that. the tensile stresses being the highest on the intake stroke, betwen TDC and mid bore. seems like thats the most important part to consider when looking at rod bolts.
 
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