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compression ratios

  • Thread starter Thread starter 67GTA-FB429
  • Start date Start date Feb 3, 2004
6

67GTA-FB429

Member
Dec 15, 2003
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16
Tri-Cities, Wa
Feb 3, 2004
#1
  • Feb 3, 2004
  • #1
When I measured my static compression last summer, after 4 compression strokes, I was getting 175-180 psi on the cylinders of my engine. How does that convert to compression ratio. I know how I did it then, but I want to know how others calculate it.

Thanks.
 

SuperDave

Early-Model Mentor
Founding Member
May 3, 2000
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0
Tacoma, WA USA
Feb 3, 2004
#2
  • Feb 3, 2004
  • #2
By "static compression" I assume (DANGEROUS here) you are talking about the cylibders' ability to hold compression over time as in a "leak down" test. Compression ratio is the numerical comparison between the VOLUME of the cylinder at the bottom and top of the stroke expressed as a ratio.

Let's clarify terms.
 
6

67GTA-FB429

Member
Dec 15, 2003
777
0
16
Tri-Cities, Wa
Feb 3, 2004
#3
  • Feb 3, 2004
  • #3
Maybe Static is the worng word.

I used a compression gauge, screwed into the spak plug hole, all spark plugs removed, carb blocked open (all per Ford Manual and Hayes), crank the motor 4 times to get a gauge reading. If barometric pressure that days was 30" Hg (or psi) and I had a reading of ~180psi, how would that convert?? I know there is a way to convert I just can't remember it.
 

Frankies65

Founding Member
Oct 10, 2001
275
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17
Bucks County PA
Feb 3, 2004
#4
  • Feb 3, 2004
  • #4
67GTA-FB429 said:
If barometric pressure that days was 30" Hg (or psi) and I had a reading of ~180psi, how would that convert?? I know there is a way to convert I just can't remember it.
Click to expand...

30"Hg = 14.7346246 psi (or divide "Hg by2.03602 to get psi)

180/14.7346246 =12.2

However, 30"hg is an absolute pressure while 180 psi is gauge pressure (if I remember my science classes properly). You would have to add 1 atmosphere to your gauge reading to make it correct, or:

(180 + 14.7346246) / 14.7346246 = 13.2

I know that most people calculate a theoretical CR, but it seems to me that an actual measurement would be a better gauge (no pun intended).

Hopefully the experts will chime in on whether it is a valid statement. I'm just doing the math.

Frank
 
J

John Z

Founding Member
Sep 21, 2000
581
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0
Morgantown WV
Feb 3, 2004
#5
  • Feb 3, 2004
  • #5
Frankies65 said:
30"Hg = 14.7346246 psi (or divide "Hg by2.03602 to get psi)

180/14.7346246 =12.2

However, 30"hg is an absolute pressure while 180 psi is gauge pressure (if I remember my science classes properly). You would have to add 1 atmosphere to your gauge reading to make it correct, or:

(180 + 14.7346246) / 14.7346246 = 13.2

I know that most people calculate a theoretical CR, but it seems to me that an actual measurement would be a better gauge (no pun intended).

Hopefully the experts will chime in on whether it is a valid statement. I'm just doing the math.

Frank
Click to expand...

I'm not sure on this but I recall this conversion of gauge reading to compression ratio is very approximate because the valves do not open and close precisely with the stroke of the pistion. In other words, you can change the gauge measured compression by changing the cam in the engine.
 

Route666

Active Member
Aug 16, 2003
1,652
6
39
Brisbane, Australia
Feb 3, 2004
#6
  • Feb 3, 2004
  • #6
Agreed, thus the static compression ratio is what he measured. I've always known static to be just that, the pressure a cylinder can create when cranked slowly, without the effects of overlap, etc.
 

Route666

Active Member
Aug 16, 2003
1,652
6
39
Brisbane, Australia
Feb 3, 2004
#7
  • Feb 3, 2004
  • #7
Frankies65 said:
30"Hg = 14.7346246 psi (or divide "Hg by2.03602 to get psi)

180/14.7346246 =12.2

However, 30"hg is an absolute pressure while 180 psi is gauge pressure (if I remember my science classes properly). You would have to add 1 atmosphere to your gauge reading to make it correct, or:

(180 + 14.7346246) / 14.7346246 = 13.2

I know that most people calculate a theoretical CR, but it seems to me that an actual measurement would be a better gauge (no pun intended).

Hopefully the experts will chime in on whether it is a valid statement. I'm just doing the math.

Frank
Click to expand...

When the gauge reads zero, there is equal pressure on both sides, ie: 14.7psi outside and inside the engine. After it's cranked, the inside reads 180psi, this includes the 14.7psi that was already in there, so there is no need to add it.
 
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