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The truth about TDC... help

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Old Oct 9, 2002 | 06:40 PM
  #1  
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The truth about TDC... help

Ok this seems to be a very confusing topic for everyone, especially me. So lets put it to rest.

Here's what i dont get.. Is TDC at the END of the compression stroke when the piston is at the "highest" point, or is it before the compression stroke actually begins.

Here's what confuses me. They say to find TDC, turn it over until you feel air on the #1 cylinder (which would be compressoin stroke), stick in a piston stop until it hits, then rotate motor backwards until it hits, then midpoint is TDC. This midpoint seems like it would be RIGHT before the compression stroke begins. But everywhere else i read looks like its at the very END of the compression stroke.

Also, if it is at the very end of the compression stroke, you would think both valves would be closed for a while before you hit TDC. If its at the beginning you would think the intake valve had just closed.

Someone help im driving myself nuts.
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Old Oct 9, 2002 | 06:46 PM
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It is at the end of the compression stroke.

Compression occurs as the piston approaches TDC. The method you refer to finds a point some small distance away from TDC, then another point equally distant from TDC on teh other side; exactly in between = TDC.

The intake valve closes somewhere during the first part of the compression stroke. The more cam you have, and the more retarded the cam is, the later during the compression stroke it remains open. That's what makes an engine with a big cam "lope".
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Old Oct 9, 2002 | 07:58 PM
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From: houston tx
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That helps a lot, but i dont understand how the piston stop method works. If you inserted the piston stop during the compression stroke it seems like it would hit the stop on its way to TDC, then when you crank it backwards it wouldnt hit it until near the beginning of the intake stroke. Thus TDC would be at the VERY beginning of the compression stroke instead of the end.

Someone staighten me out!
Thanks
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Old Oct 9, 2002 | 08:20 PM
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You make it hit the stop on the compression stroke; then you rotate it backwards until it hits it again (which will actually be just after the beginning of the exhaust stroke); the center of the "dead zone" where the piston can't get to is TDC. As far as the piston is concerned, there is no difference between the TDC at the end of compression & beginning of the power stroke, and the end of exhaust and beginning of intake.
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Old Oct 9, 2002 | 08:28 PM
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From: houston tx
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I know i'm retarded because I still dont get it.

If it hits at the end of the compression stroke, then you rotate it backwards to the beginning of the exhaust stroke, the piston will travel from all the way "up" in the bore, to all the way "Down" then back up again until it hits. if you took the middle point of that, wouldnt it be all the way down??

:-(

shane
p.s.
thanks for your patience
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Old Oct 9, 2002 | 08:32 PM
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In the cycle of events the piston is at TDC twice. Once at the end of the compression stroke and beginning of the power stroke. The second is at the end of the exhaust stroke and the beginning of the intake stroke.

At the beginning of the intake stroke the piston is near BDC.

TDC is when the piston is at its most upward position. When using the piston stop method TCD will be half way between the two marks---Half way between the shortest distance. If you find a point half way around the long distance you have found BDC.
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Old Oct 9, 2002 | 08:39 PM
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From: houston tx
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Ok one last quesiton and i think i've got it.

How can the piston be at TDC, the uppermost point, with a piston stop screwed in. Seems like it would interfere!
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Old Oct 9, 2002 | 08:47 PM
  #8  
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Try drawing it on apiece of paper. If i had a scanner i'd show you the one i just did. But trust me, it works.

Or, if you're mathematically inclined, do it like this. Imagine your dillemma of spinning the engine all the way up to the stop, then all the way back around to hit it again. Then imagine that you can extend your piston stop straight down, like in a, say, 426 hemi. So you keep winding it in. As you do this, obviously you have to crank it around less and less. Eventually you reduce to the base case where you can't move it at all, which would be..... BDC, which is exactly on the other side of....TDC.
If you imagine that you still have a tiny bit of movement left, lets call that the limit of a small angle such that you still have 2 segments. Well, TDC would be exactly the midpoint between those 2 lines. Just like if the angle was bigger.
Your mistake is you were thinking of the midpoint of the wrong half of the circle. You get a B for showing your work, but missing the solution is kinda funny.

Sorry, long day at work, bwahahaha

Hey, you ever take your girlfriends off that page?
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Old Oct 9, 2002 | 09:01 PM
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Originally posted by ShaneBuss2
Ok one last quesiton and i think i've got it.

How can the piston be at TDC, the uppermost point, with a piston stop screwed in. Seems like it would interfere!
You will have to remove the piston stop to position the piston at TDC. The piston stop is only used to find two points equal distant from TDC then the stop is removed.
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Old Oct 9, 2002 | 09:40 PM
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phew!
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Old Oct 9, 2002 | 11:56 PM
  #11  
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From: houston tx
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Originally posted by Lionsden
You will have to remove the piston stop to position the piston at TDC. The piston stop is only used to find two points equal distant from TDC then the stop is removed.
is this true? if so that makes sense the way i'm thinking about it. I just dont see how a piston stop can be screwed into the hole without infereing if a piston is at TDC. ANd obviously if you turn the engine one way, then the other, and the middle is TDC, at some point it crosses TDC.

Ed i see how a piston stop would work to find BDC, but thats becase at BDC the piston would not interfere with the piston stop.

I appreciate everyones help on this, and i'm glad yall dont know HOW mathmathetically inclined i am, but that would be even more embarassing. ;-)

And X's did get removed when new girlfriend found the links. :-(
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Old Oct 10, 2002 | 12:43 AM
  #12  
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but it doesn't cross TDC. the piston stop stops the piston from travelling upwards in the cylinder. the stop prevents it from hitting TDC. you turn the crank until the piston hits the piston stop, which prevents it from moving any further upwards.you mark a spot on the balancer. the crank is then turned in the opposite direction, so the piston goes back down, passes BDC, and then comes back up. when it hits the piston stop, another mark is made on the balancer. the distance between these two marks represents the distance the piston was not allowed to travel, because of the piston stop. and the very center of that distance, the point in the middle of those marks, is TDC. i hope that cleared it up at least a little bit for you
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Old Oct 10, 2002 | 01:18 AM
  #13  
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I think I understand the confusion here. Strange minds think alike

What you're saying is true, that once you reverse the rotation of the crank to find the other position of where the piston hits the stop, then the intake valve would then be open again, right?

I drew up a really crappy pic to use as a reference and show you how to avoid that.

Spin the crank clockwise on the comp. stroke until it hits the stop. Then spin it counter clockwise until it hits the stop again. Mark on the "degree wheel" where the pointer was on each of the two stops.

Let's use a 360* wheel for example, and the pointer stops at 2* clockwise and 318* counter clockwise. First take the higher number and subtract it from 360*. (360 - 318= 42*) Next add the lower number to that answer (42+2=44*) Then divide that number by 2 (44 / 2= 22*) Position the degree wheel so that the clockwise stop rests on 22*. Then check it by rotating the crank counter clockwise until is stops on (360 - 22= 338*) That way the amount of degrees is even on both sides of 0*. Then after you pull the piston stop out, rotate clockwise on the comp. stroke until the pointer rests at 0* on the degree wheel.

Then you are at "TRUE TDC"

AJ
Attached Thumbnails The truth about TDC... help-tdc.jpg  
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Old Oct 10, 2002 | 05:54 AM
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If it hits at the end of the compression stroke, then you rotate it backwards to the beginning of the exhaust stroke, the piston will travel from all the way "up" in the bore, to all the way "Down" then back up again until it hits. if you took the middle point of that, wouldnt it be all the way down??
No... the middle of the "zone" that the piston cannot get to, because it hits the stop on either side, is TDC. BCD is the middle of the zone that the piston can travel through.

Imagine that you put the piston at BDC, and install a stop somewhere up at the top of the cyl. Now raise the piston by rotating the crank one way; it will do something when it gets to the stop... it will stop. OK, now you mark the balancer at that point. Now you rotate the crank from that point in the opposite direction; the piston will go all the way back down, then as you continue to rotate it it will start going back up. Eventually it will hit the stop again, and .... stop. Now you mark the balancer at this point.

TDC is exactly halfway between the marks, in the "zone" that it could not travel through with the the stop in place. The center of the "zone" that it could and did travel through is BDC. But even that is perfectly useful information: if you know exactly where BDC is, then TDC is exactly 180° (½ revolution) from there.

Also, let's get past all this confusion about which stroke is where. Go back to the basics of how a heat engine using internal combustion operates. If you look at one cylinder through one complete engine cycle, there will be 2 complete rotations of the crank; 2 instances of TDC; and 2 instances of BDC. Start at the firing point, which is just before TDC. The cylinder fires, combustion occurs, and the hot gas reaches high pressure and tries to expand. This pressure pushes on the piston, and as a result the piston travels down, all the way to BDC, for ½ revolution of the crank, being pushed by the pressure of the hot gas, and allowing the gas to cool (transferring its heat energy into mechanical energy). The piston reaches BDC 180°, or ½ revolution, after firing at TDC; somewhere near that point, the exhaust valve opens, and the spent combustion by-products begin to escape. The piston reaches BDC, the crank continues to turn, the piston goes back up, and pushes all the exhaust out of the cylinder. Presently the piston reaches TDC again; near this point, the exhaust valve closes, and the intake valve opens. The crank has now made one complete revolution. The timing mark will again be aligned with the pointer, but no firing will occur this time. The piston passes through TDC again; draws in fresh air/fuel mixture through the intake valve; and as the crank continues to rotate, eventually reaches BDC the second time in the cycle. Somewhere near BDC, the intake valve closes, and the piston passes through BDC. We are now 1½ complete crank revolutions farther down the road. As the crank piston travels up, the fresh charge of A/F is compressed, preparing it for combustion; as the piston approaches TDC for the second time in this cycle, the mixture is ignited. The crank has now made 2 complete revolutions, the camshaft and distributor only one.

Got it?
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Old Oct 10, 2002 | 08:23 AM
  #15  
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From: houston tx
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Axle/Gears: 3.73 posi, stock rear
Thanks for your help, it makes sense now.

They only thing that i think i was really confused about is how the piston stop could be screwed in at TDC.

Since it can't, makes perfect sense!

Thanks again,
Shane

p.s. sweet drawing
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Old Oct 10, 2002 | 11:13 AM
  #16  
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piston stop

Does a piston stop just a long bolt that goes in the #1 plug hole?
I did a TDC on a car with a metal rod... yes i know it was not accurate but worked, so i'm interested in a piston stop, because i need to do it on my car
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Old Oct 10, 2002 | 11:50 AM
  #17  
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Why doesnt someone just define it? TDC= TOP dead center.

I think that pretty much covers it.
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Old Oct 11, 2002 | 02:35 PM
  #18  
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From: Santiago, CHILE
Car: 1986 Pontiac Firebird Trans Am
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Now,

About the "TDC" for valves:

Does the intake valve must be fully open exactly at middle of the intake stroke (1/2 piston stroke) = 90º crankshaft??? Or somewhere between 0 $ 180º?? How can I review if the camshaft timing mark is O.K.?

Other topic: If I turns "mannualy" the engine with the start key in "ON" and the # 1 spark plug into its wire doing contact with an engine metal surface, the spark plug will be ignited at the programated timing and I'll can see it and compare it with the right timing at TDC, valve position and timing marks, O.K.? I'm rigth?.

Regards,

Denis V.
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Old Oct 12, 2002 | 09:06 AM
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That point is one of the cam specs that get tossed around, but people don't understand...

The "normal" or "straight-up" install point for most aftermarket cams is with the tip of the intake lobe (peak opening) at 106° after TDC.

On your other question, theoretically, yes. But when the engine is running, all the slack in everything is fully taken up, which is very hard to do just turning it by hand; as an example one of the things that happens is the distributor shaft gets pushed up when the engine is running, which changes the timing (later) because of the spiral cut of the gears.

You can do better by turning the crank in its direction of rotation until it reaches the firing point you want, and setting the dist by pulling up on the shaft and rotating the shaft CCW against the gear as hard as you can, and adjusting the dist body until the "teeth" on the starwheel align.
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