Remove and Torque crank bolt, finally an easy method.

[Off Road SHO]

Ha! You think you had it bad, Perry? My trig tables were on Papayrus scrolls.

Tom

 

[93rev2sev]

I did, back in 1968

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

It's amazing how technology has advanced in the last 50 years. It's exciting to think about how it will advance in the NEXT 50 years, but I think you've seen the largest explosion of technology that has ever been or will ever be.

 

[intimdatr]

As far Sin Cos and Tan tables being on paper, they are still on charts of paper for us in the back of our Calc. Book.

 

[Airborne]

As far Sin Cos and Tan tables being on paper, they are still on charts of paper for us in the back of our Calc. Book.

They are back there for quick reference. You will also find measurement conversion chartsn etc. The quick reference help you with your calculus so you don't have to pull your high school books out of storage or search google.

 

[intimdatr]

Id imagine (other than my nice calculator) or math vs your math days havent changed much other than in the very upper levels.

 

[RonPorter]

It's amazing how technology has advanced in the last 50 years. It's exciting to think about how it will advance in the NEXT 50 years, but I think you've seen the largest explosion of technology that has ever been or will ever be.

Remember that, around 100 years ago, they were discussed closing the Patent Office, as they felt that everything had already been invented.....

 

[pjtoledo]

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.

in one of my first posts I agreed with the 1523.8 number, and still do.

the rest of your post is agreeing with me, I think. I'm not certain which included angle you are referring to.

restating what started the debate: I disagree that applying 200ftlbs to the crank, which is at 5 BTDC, will result in 133 lbs of force at the top of the piston.

I maintain that for a constant amount of torque applied to the crank, the force at the top of the piston will increase as the crank rotates forcing the piston towards top dead center.
note: this is referring to the crank swing being in the upper half only.
another note: if the crank is at EXACTLY top dead center, any force applied to the top of the piston will result in ZERO rotational force on the crank.
rediculous note: for a really good time, throw in offset wrist pins or bore centerlines.



I've lots of engine parts, a degree wheel, weights, and a torque wrench. See ya soon.

 

[Airborne]

I think it will help if I try to explain the rest of the forces in the free body diagram that I left out. Thanks for giving me the opportunity to explain a little more. OK, we agree that the force at the connecting rod lower bearing is 1523.8 when 200 ft-lbs is placed at the crank bolt. According to my free body diagram, at 5 degrees BTDC the resultant force in the y-axis is 133 lbs. What I did not include was the resultant force in the x-axis which is 1518 lbs. What I also did not include was the resultant force at the piston/piston pin which is 133 in the y-axis and 1518 in the x-axis. The 1518 lbs in the x-axis is directed into the cylinder wall and the 133 in the y-axis in directed to the top of the piston. So, applying Newton’s law, 1518 in the x-axis and 133 in the y-axis at 5 degrees BTDC will result in 200 ft-lbs at the crank bolt minus friction.

"" Originally Posted by pjtoledo

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? ""

Actually, this is a good example. The anchor attached to the wall has a resultant force. If we know the angle of the cable, we can calculate the force in both the x and y-axis. Because of the anchor design, the wall will receive forces (in tension) in both x and y-axis. Now imagine the anchor is n i-beam/roller bearing design and separate mechanism like a column close to the wall but supported by the floor foundation supports the cable and weight in the y-axis. Now the wall only receives the resultant force in the x-axis and the floor foundation receives the force in the y-axis.
What is happening in regard to forces of my free body diagram is similar to the above example. The lock up tool is not anchored to the piston; therefore it does not receive the force in the x-axis generated by the 200ft-lbs of torque. That force is translated to the cylinder wall in the x-axis.
Hope this helps.

 

[93rev2sev]

I've lots of engine parts, a degree wheel, weights, and a torque wrench. See ya soon.

You're gonna say "Bah, humbug" before the end.

 

[pjtoledo]

You're gonna say "Bah, humbug" before the end.

The reason I AM saying "Bah humbug" is because I didn't bet anybody.

You guys should definitely take another look.

Reconsider the example of the sign hanging from the cable, take a step back and look at the overall picture, go easy on the calculations as this has morphed into a classic "turned a mole hill into a mountain".

As an FYI, about that hanging sign, it is 1600 Newtons and is hanging from the center of a cable and pulling the cable down causing each segment of the cable to form a 25 degree angle from horizontal.
The tension on EACH SEGMENT is 1900 Newtons, significantly more than the sign itself.
I lifted that from page 115 of a physics book, the numbers are correct.




Back to the crank torque vs piston pressure. A clearer understanding of why the force goes up instead of sideways can be represented by a drawing of the parts and where they move to.


the crank is at 5 degrees BTDC.

Draw, starting with the crank center point, and the 80mm diameter circle the rod journal makes.

draw a line straight up from the crank centerpoint going about 147mm past the circle and end it at the wrist pin, thats the TDC centerline. that represents both the crank throw and rod at TDC.

draw the 5 degree BTDC line from the crank center out to the journal. it will intersect the circle a bit more than 1/8" away from the TDC centerline. That is the crank throw.

Draw the rod from the wrist pin to the rod journal at the 5 degree point.

you should now have a 167mm tall thin triangle, bottom angle 5 degrees, top angle 1+ degrees, middle angle about 174.

follow me so far?

Here comes the fun part, lock the piston and wrist pin in place while the crank is still at 5 BTDC, then cannot move up or down.

at this point the lower rod center and the crank throw are at the same identical position, but they are now disconnected.

grab your compass, set the point on the wrist pin (which is locked in place vertically) and the pencil at the rod journal and draw an arc across the TDC line simulating a swing of the rod, (remembering the rod and crank journal are still disconnected).

This arc will cross the TDC line a few thousandths below/inside the crank circle. its very close, but is inside the circle.

still with me?

we can unlock the wrist pin now

the difference between the crank circle and the "rod swing arc" on the TDC line is the distance the piston will travel up as the crank rotates from 5 BTDC to TDC.


the segment of the (1) crank circle between 5 BTDC and TDC, combined with the corresponding (2) "rod swing arc" , and the short segment of the (3) TDC line between them form a small quasi-triangle. (2 arcs and a straight line, close enough)

think of that quasi-triangle as a wedge that is forced (with about 1500 lbs)between the crank and the wrist pin to move the wrist pin up to TDC.

about that quasi-triangle....because of the arcs its mechanican advantage increases as it is driven in, nearest the TDC line the arcs are closest to parallel.



obviously this is a generalization, totally ignores friction.
Comments are welcome.

be warned: I already determined how about much crank torque is required to lift 50 lbs with the piston 1/8" below deck, no wrenches were needed.


Perry


Hey Off Road Sho, got anything for a headache?

 

[Off Road SHO]

Yeah, but she's busy right now, taking care of mine


Tom

 

[93rev2sev]

I'm going to give an example that may or may not clear this up but first, I've got to give props to Perry for sticking with this question.

Your hip is a wristpin, torso a piston. Your Femur is the connecting rod, your knee is a rod journal and your shin is the crank throw. You're standing with your back to a wall.

Bend your knee 5° and put as much weight on your shoulders as you can hold. Now bend it more. Yep, you just blew out a wristpin.

Hope this helps.

 

[Airborne]

This is a perfect application of the saying “You can’t teach an old dog new tricks”, and I also have to give props to Perry for sticking it out. And I am going to stop trying to teach, but I do have 2 suggestions if you want to pursue further:
1. Put your Physics book away and pull out your Statics book.
2. If after you pull out you Statics book and you are still having trouble “seeing” the forces in the Free Body Diagram, take the Free Body Diagram to your local Community College and find the Statics Instructor. Face to face always works the best, so much is lost over the internet. Most professors are more than willing to discuss their subject even with someone not taking their class.

 

[pjtoledo]

I'm going to give an example that may or may not clear this up but first, I've got to give props to Perry for sticking with this question.

Your hip is a wristpin, torso a piston. Your Femur is the connecting rod, your knee is a rod journal and your shin is the crank throw. You're standing with your back to a wall.

Bend your knee 5° and put as much weight on your shoulders as you can hold. Now bend it more. Yep, you just blew out a wristpin.

Hope this helps.

You don't seem to realize you are agreeing with me.
I have had back surgery 3 times.
I'm still married so there is little chance of blowing out a wrist pin.
Have you ever taken a good look at the steam locomotives at Greenfield Village?

 

[pjtoledo]

This is a perfect application of the saying “You can’t teach an old dog new tricks”, and I also have to give props to Perry for sticking it out. And I am going to stop trying to teach, but I do have 2 suggestions if you want to pursue further:
1. Put your Physics book away and pull out your Statics book.
2. If after you pull out you Statics book and you are still having trouble “seeing” the forces in the Free Body Diagram, take the Free Body Diagram to your local Community College and find the Statics Instructor. Face to face always works the best, so much is lost over the internet. Most professors are more than willing to discuss their subject even with someone not taking their class.

OK, to make certain exactly what it is that we are disagreeing on.


After a few posts, you determined that applying 200 lbs ft to a crankshaft, whose rod journal was positioned at 5 degrees before top dead center, would cause the piston to apply 133 lbs of force to a stop tool.

Is that correct?


I maintain the force of the piston against the stop tool will exceed 133 lbs by a significant amount.

Is that correct? ( scratch correct, insert: is that what you disagree with)

I am not contesting other static forces, their directions or magnitudes. I know they exist and things break if structures cannot withstand them.


I have a working model with which to take measurements.
I'll stack approx 135 lbs on the deck surface, use spacers of various thickness on top of the piston to stop the crank at a few different angles before top dead center. Then note the torque needed to raise the weights off the deck.

Is that a suitable test procedure?
Will I get different results if I substitute a rigid bar in place of the weights?
(I have the sneeky suspicion that "static" may bite my **** there)
I won't be able to measure the pressure between the piston/spacers and the rigid bar.

Preliminary testing last night went as such:

with a 1/8" spacer on top of the piston and aligned with the deck, which put the crank a few degrees before top dead center, and 50 lbs stacked on the deck I was able to rotate the crank from well before TDC, thru TDC (lifted the weights), to well after TDC using only my hand gripping the crank snout. No tools whatsoever.
I'll increase the weight, obtain some data points and post them up sometime soon.

 

[intimdatr]

Lol. I like my rope.

 

[pjtoledo]

did some tinkering down in the basement. a final number has been determined.

Anybody care to duplicate the "lab" results? actual testing, no calculations allowed.

I only had 1 piston installed, kept friction to a minimum.



Perry

 

[93rev2sev]

You need one of these for a static experiment.

 

[Airborne]

did some tinkering down in the basement. a final number has been determined.

Anybody care to duplicate the "lab" results? actual testing, no calculations allowed.

I only had 1 piston installed, kept friction to a minimum.



Perry

I conducted my experiment last January when I installed the lock up tool in my engine and placed 200 ft-lbs of torque on the crank bolt and no damage to the cylinder head or the lock up tool and have put 10k miles since. And twice since then on other SHO's while helping others do front 60k.

 

[pjtoledo]

I conducted my experiment last January when I installed the lock up tool in my engine and placed 200 ft-lbs of torque on the crank bolt and no damage to the cylinder head or the lock up tool and have put 10k miles since. And twice since then on other SHO's while helping others do front 60k.

I never said the tool wouldn't work.
I am aware that if the threads were unwound they would be of considerable area, thus strength also.
Some plastics have considerable compression characteristics.
The tools have been sold for years, I have no doubts that they work.


The issue I dispute is the resultant 133 lbs of force at the piston top when 200 lbs-ft is applied to the crank, at 5 degrees BTC.


Here is my test procedure.

3.2 block with only 1 piston, no rings. kept friction to a minimum.
Start with the piston at BDC, suitable cylindrical standoff (big soup can) to clear deck by about an inch.
Add a constant force of 140 lbs on top of the standoff, 50 x 2, 15 x 2, 10 x 1 exercise weights. A little difficult to stabilize, but it can be done.

Applied torque wrench to crank bolt, rotated crank to run piston from BDC to TDC.
The pull started at BDC with virtually no resistance, increased rapidly right around the half way point, maxed out about 3/4 of the way up, then tapered off the virtually nothing at TDC.
To my surprise the maximum resistance was with the piston about 3/4 of the way up, I had expected it to be at the half way point.

After trial and error, and a slow steady hand, the torque at the maximum resistance was less than 25 lbs-ft. 24 would click, 25 did not click.

Yes that means with a wrench set to 25 lbs-ft I was able to crank the piston from BDC to TDC while 140 lbs was balanced on top of the piston/standoff combination.
The torque between 177 and 183 ( the bottom of the crank swing) was so low I could rotate the crank with my hands.
(Don't try it at the top without safety stops)

5 degrees BTDC puts the piston within a few thousands of the top. I have a dial indicator, may be I'll get some actual measurements this weekend.

I would expect the stop tool to work best at the point of max resistance.

Anybody close to Toledo have dial type wrench? Would make plotting a graph a lot easier.

I look forward to someone trying this and posting their results.


On a side note, it would be interesting to try the other side of the crank stroke. I always wondered if these blocks had an offset cylinder bore, and what the effects would be.