Remove and Torque crank bolt, finally an easy method.

[auto.guru]

auto.guru - Pay attention to the mileage in sdpatt's sig and realize he's the original owner.

actually it just says
Helping to keep the SHO on the road,
Scott
1991, 425K miles, and rolling...
In-car Track Video: Laps at TWS


so... idk how im wrong on this is missed something? anyway...
i barely passed math back in highschool for a reason. stop bringing this stuff up. my lord... lol

 

[SHOmethewayhome]

he was implying that SDPatt has over 425 thousand original miles on his one owner SHO without the harmonic balancer separation issue occuring.

i've had it happen 3 times in the last 10 years on ford Windsor blocks (but these were all truck motors that had seen 200K of hard life and had never been replaced and 2 of them lived in very hot, very dry enviroments and that probably compromised the rubber in the harmonic balancer.) I also usually just block the ring gear and/or put the car in gear and step on the brakes when possible.

that tool is pretty sweet though.

you know the fastest way to tell someone's an engineer? they'll tell you.

 

[Airborne]

you know the fastest way to tell someone's an engineer? they'll tell you.

LOL, I have always been able to laugh at myself. Ahh, did I do it again?

 

[Off Road SHO]

My head hurts. No more math allowed in this thread or I will turn it around and march you guys right back....Oh, wait a minute, wrong nightmare.

Carry on.

Tom

 

[93rev2sev]

I set out to debunk this but...

Let’s talk about the forces on the spark plug threads. Normal production engines produce between 800-1000 psi during the compression stroke. High compression race engines produce about 1500 psi. Top Fuel produces well over 2000 psi on the same 14mm plug and aluminum threads. So how much force is exerted on the plug forcing out of the cylinder a hundred times a minute for hours?
Area of the plug:
A = Πr2 = 3.14(0.28) 2 = 0.2463 in2
Multiply 0.2463 in2 by 1500 psi and you get: 370 pounds of dynamic force exerted on the face of the plug trying to force it out of the cylinder.
How much force are we putting on the threads by using the Engine Lock-Up Tool?
Let’s assume you wanted to place 200 ft lbs of torque on your crank bolt. I used 85° because the Lock-Up Tool stopped the crank from moving at about 5° before TCD. That would equate to 133 lbs of static force exerted on the plug tool trying to force it out of the cylinder head. See diagram below.

 

[pjtoledo]

so, according to Newton's 3rd law of motion: if the piston pushes the stop plug with X force, then the stop plug also pushes the piston with the same force X , correct?

I can't wait to tell some engineers that you can set an engine at 5 degrees before top dead center, push down on the piston with about 133 lbs, and get over 1500 lbs of force exerted on the rod journal trying to rotate the crank.




perhaps I can offer an example that may help us understand.


conside a weight hanging from a cable strung between 2 very solid, immoveable supports.
the weight (1600 newtons) which is supported at a single point at the exact middle of the cable, causes tension on the cable. That tension pulls the cable down, resulting a 25 degree angle from horizontal at each end.

How much tension is on each cable segment, one from the sign to left anchor point, the other from the sign to the right anchor point?


Back to Newton's 3rd law, A pushing B equals B pushing A.


that means we can substitute the cables with solid rods, and push UP on the sign with a force of 1600 N and get the pressure on the rods to equal the tension on the cables. Correct?

substituting again, put the connecting rod on the left, (or right) and the crank throw on the right (or left), and the force of 1523.8 lbs substituted for the sign, still pushing up.

Calculate the first answer using the 25 degree angles. Then run it again using 5 degree angles to simulate 5 degrees before top dead center.



(Note: to simplify things I stroked the crank out to equal the rod, egos may have been damaged, but the theory remains the same.)

(Another note: I'm pushing straight up on the 1523.8 lb sign, there would be a slightly different answer if the angle of the segment opposite the 5 degree angle were included. That would cause more tension on one side, less on the other.)

Perry

 

[Airborne]

You should be commended for doing this research and investigation. But you are assuming that all force is being transmitted to the y-axis and not visualizing the force in the x-axis. This is common for someone not formally trained in Statics and Dynamics, but still you should be commended. I didn’t show the force exerted in the x-axis because it did not apply to my point, but is still there, it is transferred between the bearings and the cylinder wall. And yes, when applying Newton’s law, both the forces in the x-axis and y-axis would translate to 200 ft-lbs of torque, at the crank, minus friction. And yes, at 25 degrees the force in the y-axis would be significant and much less in the x-axis. That is why this tool stops the crank at 5 BTDC and not 25 BTDC.

 

[Shovert]

The lockup tool should work. Reason I say. When they pulled my motor they damaged the pulley. So needed another pulley. So off to the Pull-a-Part and found a motor. I removed a sparkplug. Rolled in some rope I brought with me into the cylinder. Then proceded to turn the bolt. 1/2 strong arm and pipe laying in the yard and bolt was loosen, backed it up, removed the rope and I removed the bolt and pulley.
I am unsure if someone wants to use this method on a good motor but works. Maurice

 

[pjtoledo]

You should be commended for doing this research and investigation. But you are assuming that all force is being transmitted to the y-axis and not visualizing the force in the x-axis. This is common for someone not formally trained in Statics and Dynamics, but still you should be commended. I didn’t show the force exerted in the x-axis because it did not apply to my point, but is still there, it is transferred between the bearings and the cylinder wall. And yes, when applying Newton’s law, both the forces in the x-axis and y-axis would translate to 200 ft-lbs of torque, at the crank, minus friction. And yes, at 25 degrees the force in the y-axis would be significant and much less in the x-axis. That is why this tool stops the crank at 5 BTDC and not 25 BTDC.

I'll do some more thinking and get back to 'ya. :wave: (I'm off on a quest to find my trusty 'ol slide rule now.)

 

[EL SHO]

So if one were to try to torque the crank bolt in an ATX car with the front wheels on the ground and with the trans in 1st gear or Parking, would that allow for the cank bolt to be torqued up to 200 ft-lbs ?

 

[Airborne]

The lockup tool should work. Reason I say. When they pulled my motor they damaged the pulley. So needed another pulley. So off to the Pull-a-Part and found a motor. I removed a sparkplug. Rolled in some rope I brought with me into the cylinder. Then proceded to turn the bolt. 1/2 strong arm and pipe laying in the yard and bolt was loosen, backed it up, removed the rope and I removed the bolt and pulley.
I am unsure if someone wants to use this method on a good motor but works. Maurice

Back in the day we used the rope trick on 2 stroke (no valves) dirt bikes.

 

[Off Road SHO]

This tool was invented by a mechanic who was tired of all the time it took to do the rope method, not to mention the aggravation when he had the piston on the wrong stroke. Been there done that.

Tom

 

[93rev2sev]

The math is 9th grade algebra.

I only remember spending about a day on vector math, but I'm pretty sure it was in pre-calc. Which, I think, came after geometry and level 2 algebra.

 

[waffles]

Question: Could a guy just put the car in 5th, put the parking brake on, chalk the wheels and go to town with a long breaker bar and a cheater?

 

[93rev2sev]

Question: Could a guy just put the car in 5th, put the parking brake on, chalk the wheels and go to town with a long breaker bar and a cheater?

Only if you HAVE 5th gear. This is most useful for ATX cars.

 

[93rev2sev]

PJToledo,

I think you're confusing yourself. 200 foot pounds on the crank bolt equates to the rod journal travelling tangentially (not up the bore) with a much larger force (simple mechanical advantage). Airborne worked out about 1550 lbs of force and I won't argue that. The crank throw is about 1.55 inches. (12\1.55)*200= 1548.

This force is tangential to the circle made by the crank's rod journal and it won't change as you approach TDC. As long as you apply 200 ft lbs to the crank bolt, the crank journal (1.55 inches from the crank centerline) will have 1548 lbs of tangential force behind it.

Now, how much of that force is applied to the piston? That depends on the included angle. If the included angle was 90 degrees, then almost ALL of it would be applied to the piston.

If the included angle was ZERO, then NONE of the force would be applied to the piston and since the piston isn't moving, contact with the engine lock up tool would have no effect at stopping the assembly from rotating.



Another way to look at this problem is to imagine the opposite. You're wanting to rotate the engine by pushing down on a piston. the best place to do that would be 90 degrees Before or after TDC as most of the force pushing down at that point would translate into engine rotation at a ratio of about 1:1.

As the pistons approach the bottom of the bore, less and less of your pushing down on it will translate into crank rotation...so you'll find another cylinder with the piston about halfway up and push on that one.

I'm 100% sure that Airborne's math is acceptable.

so, according to Newton's 3rd law of motion: if the piston pushes the stop plug with X force, then the stop plug also pushes the piston with the same force X , correct?

I can't wait to tell some engineers that you can set an engine at 5 degrees before top dead center, push down on the piston with about 133 lbs, and get over 1500 lbs of force exerted on the rod journal trying to rotate the crank.




perhaps I can offer an example that may help us understand.


conside a weight hanging from a cable strung between 2 very solid, immoveable supports.
the weight (1600 newtons) which is supported at a single point at the exact middle of the cable, causes tension on the cable. That tension pulls the cable down, resulting a 25 degree angle from horizontal at each end.

How much tension is on each cable segment, one from the sign to left anchor point, the other from the sign to the right anchor point?


Back to Newton's 3rd law, A pushing B equals B pushing A.


that means we can substitute the cables with solid rods, and push UP on the sign with a force of 1600 N and get the pressure on the rods to equal the tension on the cables. Correct?

substituting again, put the connecting rod on the left, (or right) and the crank throw on the right (or left), and the force of 1523.8 lbs substituted for the sign, still pushing up.

Calculate the first answer using the 25 degree angles. Then run it again using 5 degree angles to simulate 5 degrees before top dead center.



(Note: to simplify things I stroked the crank out to equal the rod, egos may have been damaged, but the theory remains the same.)

(Another note: I'm pushing straight up on the 1523.8 lb sign, there would be a slightly different answer if the angle of the segment opposite the 5 degree angle were included. That would cause more tension on one side, less on the other.)

Perry

 

[intimdatr]

As i am currently in Physics ^ i agree with what hes said.

And this Math is Pre-Calc through Calc. 2. Seeing as Calculus was invented to understand and prove Physics. Far from simple algebra days of 9th grade

Although my vector math skills are not the best. I believe 93res2sev Math and ideals to be correct.

 

[Airborne]

Using COS, SIN and TAN to find the length of a triangle side given a known angle is NOT calculus. This is nothing different except we have a known force instead of length. 9th grade algebra plain and simple. But the concept of the free body diagram and applying the math is way past 9th grade. I think you guys are confusing the two.

 

[93rev2sev]

Using COS, SIN and TAN to find the length of a triangle side given a known angle is NOT calculus.

You're right. But lets not get too stuck on it.

This site would not be needed if every 9th grade algebra student had been REALLY paying attention...

http://hyperphysics.phy-astr.gsu.edu/hbase/vect.html

 

[pjtoledo]

You're right. But lets not get too stuck on it.

This site would not be needed if every 9th grade algebra student had been REALLY paying attention...

http://hyperphysics.phy-astr.gsu.edu/hbase/vect.html

I did, back in 1968

we used trig tables then, as there weren't any affordable or portable calculators.