Only because Stubby suggested it I'll post a method of how to determine your CR from the compression test gauge reading. I thought I might just waffle a bit and write on the run. So forgive me if it isn't eloquent.
Most will be aware that CR is a handy way of guessing the performance of an engine. For instance with high CR we run higher octane fuels, have a cam that probably opens and blows down a little earlier, thus a few degrees extra LSA, etc. But best of all it's still good for bragging rites and big noting......"my engine is running CR 12.5, what size is your dick?"
CR in isolation doesn't help when trying to figure if we are going to get det or not. That's when the effective or dynamic compression ratio comes in handy. It's simply the ratio of volume available after the intake cycle divided by the clearance volume (clearance volume is all that space above the compression ring). The theory is that while the intake valve is open there isn't much compression happening.
Suprisingly we can use this to fairly accurately determine our expected compression pressure and by extrapolation the BMEP too, which is a bonus.
So lets take a Ford 3984cc O/head cam six from the EA -EF series Falcons as our example.
What we know ( I hope I'm right here, because I'm not really a Ford type):
CR = 8.8:1
Bore diameter = 92.25
Stroke = 99.31mm
Compression 1010 kPa minimum
Camshaft advertised events 65/11 17/63
Conrod length (centreline) 153.92mm
Note: I'm using metric here, but you can use inches if you want.
1. The core calculation is to find the displacement of the piston left to travel to top dead centre (TDC) once the intake valve has closed. This is fairly easy if you know trigonometry. and can be found by using::
Effective stroke = RL- SQRT(RL^2 - (0.5 x S x Sin°)^2) + (0.5 x S x Cos°) +(0.5 x S)
where RL = Rod length = 153.92, S = stroke = 99.31mm, ° = inlet closing angle =63, ^2 = squared (raised to the power of 2)
So in a very convoluted way;
effective stroke = 153.92 - SQRT(153.92^2 - (0.5 x 99.31 x 0.891)^2) -(0.5 x 99.31 x 0.454) + (0.5 x 99.31)
=153.92 - SQRT(23691.4 - 1957.4) + 22.54 + 49.655
=153.92 - 147.42 + 22.54 + 49.655
= 78.695mm
2. From this we can work out our effective compression ratio if we know the static compression ratio, which we do in this instance = 8.8:1
DCR = (Effective stroke ratio x (CR-1)) + 1
effective stroke ratio = 78.69/99.31 = 0.792
= (0.792 x 7.8 ) +1
= 7.1776
3. then we can apply our ideal gas law and find the our expected reading on the compression gauge to see things are going well with our rings and valves:
(DCR^0.283 x atmos pressure x DCR)- atmos pressure
(7.1776^0.283 x 101.32 x 7.1776 ) - 101.32
=1169 kPa
4. But what if you want to work out what the clearance volume is (for some obscure reason) and you don't want to take the head off and start the onerous task of finding the combustion volume, ringland volumes, gasket volumes, etc
Well we take the gauge reading and work out the DCR by rearranging variables with simple algebra and get:
DCR = (1169kPa +101.32/101.32)^(1/1.283)
= 12.5377^0.77942
= 7.1775
we know from the previous calculation the effective stroke is 78.695mm
we also know that CR = (Full stroke swept volume/clearance volume) +1
so it follows that DCR =(effective swept volume/clearance volume) +1
therefore clearance volume = effective swept volume/(DCR-1)
effective swept volume = (92.25^2 x Pi x 78.65)/(4 x 1000) = 525.68 cc
so clearance volume = 525.68/6.1775 = 85cc
5. Knowing the clearance volume you can easily work out the CR value, if it's important to you.
.........................
What is the use of DCR you ask? Well it is particularly useful in predicting detonation. And none of us want det, because it has a nasty habit of causing erratic power and damage. It is also is a good predictor of BMEP, which is a good predictor of something novices like to brag about ..... engine torque.......which is really just cross product vector of power and annular velocity .
You can use DCR to select the correct fuel octane rating. And very importantly you can maximise the compression ratio for use in turbo or huffer applications, without sticking to that old chestnut that CR must be around CR 8 to prevent catastrophic failure.
How did I do? Clear as mud?
Most will be aware that CR is a handy way of guessing the performance of an engine. For instance with high CR we run higher octane fuels, have a cam that probably opens and blows down a little earlier, thus a few degrees extra LSA, etc. But best of all it's still good for bragging rites and big noting......"my engine is running CR 12.5, what size is your dick?"
CR in isolation doesn't help when trying to figure if we are going to get det or not. That's when the effective or dynamic compression ratio comes in handy. It's simply the ratio of volume available after the intake cycle divided by the clearance volume (clearance volume is all that space above the compression ring). The theory is that while the intake valve is open there isn't much compression happening.
Suprisingly we can use this to fairly accurately determine our expected compression pressure and by extrapolation the BMEP too, which is a bonus.
So lets take a Ford 3984cc O/head cam six from the EA -EF series Falcons as our example.
What we know ( I hope I'm right here, because I'm not really a Ford type):
CR = 8.8:1
Bore diameter = 92.25
Stroke = 99.31mm
Compression 1010 kPa minimum
Camshaft advertised events 65/11 17/63
Conrod length (centreline) 153.92mm
Note: I'm using metric here, but you can use inches if you want.
1. The core calculation is to find the displacement of the piston left to travel to top dead centre (TDC) once the intake valve has closed. This is fairly easy if you know trigonometry. and can be found by using::
Effective stroke = RL- SQRT(RL^2 - (0.5 x S x Sin°)^2) + (0.5 x S x Cos°) +(0.5 x S)
where RL = Rod length = 153.92, S = stroke = 99.31mm, ° = inlet closing angle =63, ^2 = squared (raised to the power of 2)
So in a very convoluted way;
effective stroke = 153.92 - SQRT(153.92^2 - (0.5 x 99.31 x 0.891)^2) -(0.5 x 99.31 x 0.454) + (0.5 x 99.31)
=153.92 - SQRT(23691.4 - 1957.4) + 22.54 + 49.655
=153.92 - 147.42 + 22.54 + 49.655
= 78.695mm
2. From this we can work out our effective compression ratio if we know the static compression ratio, which we do in this instance = 8.8:1
DCR = (Effective stroke ratio x (CR-1)) + 1
effective stroke ratio = 78.69/99.31 = 0.792
= (0.792 x 7.8 ) +1
= 7.1776
3. then we can apply our ideal gas law and find the our expected reading on the compression gauge to see things are going well with our rings and valves:
(DCR^0.283 x atmos pressure x DCR)- atmos pressure
(7.1776^0.283 x 101.32 x 7.1776 ) - 101.32
=1169 kPa
4. But what if you want to work out what the clearance volume is (for some obscure reason) and you don't want to take the head off and start the onerous task of finding the combustion volume, ringland volumes, gasket volumes, etc
Well we take the gauge reading and work out the DCR by rearranging variables with simple algebra and get:
DCR = (1169kPa +101.32/101.32)^(1/1.283)
= 12.5377^0.77942
= 7.1775
we know from the previous calculation the effective stroke is 78.695mm
we also know that CR = (Full stroke swept volume/clearance volume) +1
so it follows that DCR =(effective swept volume/clearance volume) +1
therefore clearance volume = effective swept volume/(DCR-1)
effective swept volume = (92.25^2 x Pi x 78.65)/(4 x 1000) = 525.68 cc
so clearance volume = 525.68/6.1775 = 85cc
5. Knowing the clearance volume you can easily work out the CR value, if it's important to you.
.........................
What is the use of DCR you ask? Well it is particularly useful in predicting detonation. And none of us want det, because it has a nasty habit of causing erratic power and damage. It is also is a good predictor of BMEP, which is a good predictor of something novices like to brag about ..... engine torque.......which is really just cross product vector of power and annular velocity .
You can use DCR to select the correct fuel octane rating. And very importantly you can maximise the compression ratio for use in turbo or huffer applications, without sticking to that old chestnut that CR must be around CR 8 to prevent catastrophic failure.
How did I do? Clear as mud?