Quote from danowat :

Anyone else just get "blah blah blah" when you read these type of threads?

Me too. It may be true that I wrote it, but if I had to read it again a couple of weeks later I'd also have to think hard to understand it. That's mostly though because it's in english and the formulae are formatted badly. Some LaTeX would help the formulae a lot.

Anyway, half an hour ago I was walking back home and had a thought about the issue. Maybe I can improve the estimation...

Let's start with a new assumed mechanism:
1. At the start of the discussion the journal is empty and contains no fluid.
2. The journal now starts to align with the oil delivery system.
3. Oil starts to flow into the journal.
4. The fail criteria is that the journal must be filled with oil up to the center of the bearing before it is shut off from the oil supply.

Let's calculate!
1. The backpressure (due to centrifugal forces) acting against the oil delivery is assumed to be constant as discussed in the last reply (0.36psi according to Tristan - actually that's the peak achieved once the journal is completely filled, but we'd like to overestimate the minimal oil pressure rather than underestimating it)
2. The two cross-sections move relative to each other at a speed of w*r. The time they intersect works out to be t = d/(w*r). At the beginning of this duration the intersecting area is infintely small, in the middle it's exactly pi*d² and at the end it's infinitely small again. Plotted over time the available area for oil transport looks like a sine-curve centred around 0.5*pi*d².
3. We neglect any time-based effects on the fluid resistance caused by the edges of the drillings and thus can handle the intersecting drillings as if they intersected with an area of 0.5*pi*d² for the whole duration of t.
This is a very bad assumption. At the end of the calculation we should at least double the calculated minimum pressure due to the neglected effects here. Or just look further down and integrate the respective equotation over time - good luck .
4. During the duration t the minimum oil pressure needs to transport a volume of V = pi*d²*(D/2) (just enough to fill the journal to the middle) through a throttle of the area 0.5*pi*d².
Thus we receive a necessary flow rate of Q = pi*d²*(D/2)/t.
6. I treat the problem as an orifice plate with incompressible fluid (Wikipedia). It's bigger diameter is d, the smaller diameter d' calculates as follows:
pi*d'² = 0.5*pi*d²
d' = sqrt(0.5*d²)
d' = d/sqrt(2)
Thus we can use the formula provided by wikipedia:
Q = pi*d²*sqrt(1/(1-(d'/d)^4)*sqrt(dp/v)
(v is density, dp difference in pressure at the orifice)
(At this point you may want to use a more sophisticated formula. This one neglects compressibility. The wikipedia link provides many formulae.)
Rearrange:
dp = Q²*v/(pi*d²*sqrt(1/(1-(d'/d)^4))²
And we finally receive the desired necessary theoretical oilpressure p as
p = Q²*v/(pi*d²*sqrt(1/(1-(d'/d)^4))² + v*w²*0.5*D² + v*g*D
Modify as desired, e.g. multiply by two for security. But please repeat the whole calculation yourself. This post should only outline a possible method of estimation.

...That was fun.

Vain

Quote from tristancliffe :

No, I didn't get to that. I'm too stupid to get that far with real problems

Approx values
v = 810kg/m³
d = don't know, as I never measured it before building the engine! But it doesn't matter in the pressure calculation anyway
D = 0.0275 (I believe, from memory, the journal is 55mm dia)
w = let's say 5000rpm, which is 523 rads/sec

Which gives 2527 Pa, which is (for my old fashioned brain; I can't do SI units of pressure and understand what they mean in the real world) 0.36psi - which is wrong.

So, either the calculation is wrong (doesn't look like it), my units are wrong (don't think so), or you are correct that the solution is a uselessly small part of the whole thing, and that overcoming the rotation isn't much of a problem and isn't the main requirement of oil pressure.

Me likes interesting topics. Thanks.

Viscosity is measured in centipoise or centistokes and totally dependant on temperature and flowrate. The Reynolds number needs to be calculated in order to determine either laminer or turbulent flow in the system. Likely laminer in your case.
Once this Reynolds number (unitless constant) is calculated then the correct formula can be applied to calculate and pressure differentials over a given system. (I like Crane for these sort of calcs)

The value you have given for v is actually the density of the oil (rho).

Also

1 Mpa = 1x 10^-5 bar

1 bar = 14.5 psi (approx).

Try getting free download of "crane flow of fluids through valves fittings and pipe" for the required formula and easy to follow fluid dynamics guide. (I used to have pdf document but can't find it right now:shrug

In general I would say that the force of rotating bearing would require negligeable pressure to overcome. Infact likely you would get suction through a venturi type effect. So the pressure you will need is totally dependent on the diameter of the annulus and the temperature of the oil.


(Excuse my spellling)


If you are struggling with Crane let me know and I'll try and help although these days my Engineering involvement is totally employed with harrassing subcontractors

Quote from Vain :

Me too. It may be true that I wrote it, but if I had to read it again a couple of weeks later I'd also have to think hard to understand it. That's mostly though because it's in english and the formulae are formatted badly. Some LaTeX would help the formulae a lot.

Anyway, half an hour ago I was walking back home and had a thought about the issue. Maybe I can improve the estimation...

Let's start with a new assumed mechanism:
1. At the start of the discussion the journal is empty and contains no fluid.
2. The journal now starts to align with the oil delivery system.
3. Oil starts to flow into the journal.
4. The fail criteria is that the journal must be filled with oil up to the center of the bearing before it is shut off from the oil supply.

Let's calculate!
1. The backpressure (due to centrifugal forces) acting against the oil delivery is assumed to be constant as discussed in the last reply (0.36psi according to Tristan - actually that's the peak achieved once the journal is completely filled, but we'd like to overestimate the minimal oil pressure rather than underestimating it)
2. The two cross-sections move relative to each other at a speed of w*r. The time they intersect works out to be t = d/(w*r). At the beginning of this duration the intersecting area is infintely small, in the middle it's exactly pi*d² and at the end it's infinitely small again. Plotted over time the available area for oil transport looks like a sine-curve centred around 0.5*pi*d².
3. We neglect any time-based effects on the fluid resistance caused by the edges of the drillings and thus can handle the intersecting drillings as if they intersected with an area of 0.5*pi*d² for the whole duration of t.
This is a very bad assumption. At the end of the calculation we should at least double the calculated minimum pressure due to the neglected effects here. Or just look further down and integrate the respective equotation over time - good luck .
4. During the duration t the minimum oil pressure needs to transport a volume of V = pi*d²*(D/2) (just enough to fill the journal to the middle) through a throttle of the area 0.5*pi*d².
Thus we receive a necessary flow rate of Q = pi*d²*(D/2)/t.
6. I treat the problem as an orifice plate with incompressible fluid (Wikipedia). It's bigger diameter is d, the smaller diameter d' calculates as follows:
pi*d'² = 0.5*pi*d²
d' = sqrt(0.5*d²)
d' = d/sqrt(2)
Thus we can use the formula provided by wikipedia:
Q = pi*d²*sqrt(1/(1-(d'/d)^4)*sqrt(dp/v)
(v is density, dp difference in pressure at the orifice)
(At this point you may want to use a more sophisticated formula. This one neglects compressibility. The wikipedia link provides many formulae.)
Rearrange:
dp = Q²*v/(pi*d²*sqrt(1/(1-(d'/d)^4))²
And we finally receive the desired necessary theoretical oilpressure p as
p = Q²*v/(pi*d²*sqrt(1/(1-(d'/d)^4))² + v*w²*0.5*D² + v*g*D
Modify as desired, e.g. multiply by two for security. But please repeat the whole calculation yourself. This post should only outline a possible method of estimation.

...That was fun.

Vain

Thanks Vain, that cleared it up nicely mate , nothing like resorting to laymans terms to help us chimps and chimpesses understand

Quote from danowat :

Thanks Vain, that cleared it up nicely mate , nothing like resorting to laymans terms to help us chimps and chimpesses understand

Frankly, I'm still completely flummoxed.

Quote from S14 DRIFT :

Frankly, I'm still completely flummoxed.

That's what he meant.

Sarcasm detector FAIL!!!

:doh:

It's National No Sarcasm day

...



:hide:

tristan, you said you are losing pressure in left handers. does your crank rotate clockwise (when looking forward from behind the car)?

Sorry for the slow reply - busy weekend.

The crank rotates clockwise when looking at the FRONT of the engine, and hence counterclockwise when looking from behind. The oil is being flicked/scraped off the crank on the left hand side mostly, collecting in the sump, and being pumped out on the left hand side with a single scavenge pump. There is no room to fit a scavenge pump on the other side, and what we have was presumably good enough for the previous owner and the F3 teams.

Quote from tristancliffe :

One for the budding engineer or mathematician...

I'm trying to calculate the minimum oil pressure required in an engine. Pretty much all oil pressure is needed for is to overcome centripetal force in the main bearings, to get into the drillings to the big end bearings. But how to calculate how much?

I know that Force = mass x angular velocity squared x radius (F=mrw^2)

But the oil pressure acts over the whole radius, from 0 (or effectively zero, but it's pretty negligable as D approaches zero), so I guess I need to integrate that - e.g. to m*w^2*(0.5*D^2) - using mw^2 as a constant.
But the mass varies over the radii too, as there isn't much oil at the middle, but there is the whole diameters worth at the extremes... So does mass need integrating too, at the same time as radius, and if so how would I do that?
We can, at least, assume w is constant...

I also tried doing it a different way, and assuming 1cc of oil in a tube of cross sectional area of 1cu/cm, and then taking F=mrw^2 at discrete radii...

But I'm not thoroughly confused!

Anybody want to help out with some simple maths (that I'm too stupid to recall)? I've not been able to find this done before on the net (but I'm sure it has been done, as it's an important calculation for real life engine design). It's not on Wikipedia either that I'm aware of

Ta!

Well firstly your equation for centripetal force appears to be wrong.
The velocity isn't angular it's tangental and is given by V not W. I thought at first that maybe it was an engineering version of the equation I'm aware of but I can't see any mention of it on any physics sites.

So:

F= (MV^2)/R rather than MV^2R as you've stated.

This would give an acceleration at any given radius of V^2/R

If you were to section up your tube of oil in to discrete sections of say 1mm and assumed the oil filled the tube completely you could then use the static fluid pressure equation (used to calculate pressure at a given water depth) of:

P= dgh

where d = density (of the oil), g= acceleration (normaly due to gravity but you could substitute in V^2/R), h = height of the column (in this case 1mm).

Calculate that for each of your discrete sections of tube and then sum them and that should give you the final pressure at the appature of your tube, ie the pressure required to be overcome to allow the oil to flow in to the tube.

I think

v and w are not the same.

v = tangital velocity = ds / dt (m/s)
w = angular velocity = d(theta) / dt (rad/s)

Angular acceleration = r x w^2

So F = ma
= m x r x w^2 = what Tristan's come up with for the angular force.

Quote from ajp71 :

v and w are not the same.

v = tangital velocity = ds / dt (m/s)
w = angular velocity = d(theta) / dt (rad/s)

Angular acceleration = r x w^2

So F = ma
= m x r x w^2 = what Tristan's come up with for the angular force.

Oh that makes it much clearer, thanks....!

Quote from ajp71 :

v and w are not the same.

Clearly.

Quote :

v = tangital velocity = ds / dt (m/s)
w = angular velocity = d(theta) / dt (rad/s)

Angular acceleration = r x w^2

So F = ma
= m x r x w^2 = what Tristan's come up with for the angular force.

The accepted equation for centripetal force is

F=MV^2/R

Using tangental velocity.

I hadn't seen the derivation of that equation using W before, so I didn't recognise it as the correct formula.

Anyway, I now realise that substuting in the relationship between angular and tangential velocity (V=WR) does indeed give the equation stated.

That's what you get for trying to use your brain when half way to the land of nod.

Quote from tristancliffe :

It's not a bad idea. I wonder if that is likely to be closer to reality than my first method, as the hydrodynamic oil wedge will be quite different to a static floating condition...

I'd love to run several engines, all at different rpm, under lateral G, gradually lowering the oil pressure until each fails. But alas I only have two engines (one in use, one a spare that we don't want to use if we don't have to), and a very limited budget. Hence finding out everything we can, from chassis behaviour and setup balance, to fuel behavious in the float chambers etc. Yes, it's all very simplified, but gradually you build up a picture of simplifiedness versus real life on-track experience (with some LFS-learnt theory too ), and come out at pretty close to the right answers.

At least, that's how it's worked over the last two years, and going from the back of the grid 2 years (and one day) ago to winning a championship 8 months ago, to instantly being on the pace of the experienced Dallara people 6 weeks ago (and hopefully again two weeks from now).

At Snetterton, during our test day, we experienced low oil pressure (I forget the actual figure, but worryingly low) in the two left hand bends - braking for the Esses and turning through Russells. This was traced to the car being underfilled, but no harm was done (phew). At the race meeting at the same circuit the pressures stayed much higher, but still a little low for comfort - only on left hand bends it seems, as though the dry sumping arrangement, which isn't easily modified, isn't good enough to maintain oil flow/pressure in left handers.

Mallory, the next race, is primarily right handers, but does have one fast left hander of any significance. Donington has a couple of left handers that might be a problem. Croft has two or three places where a problem might occur. Silverstone has one, and then back to the relatively safe Snetterton... I'd like to kill a potential problem before it does anything bad - but to know if we do actually have a problem I wanted to know what the minimum oil pressure is to ensure that the oil gets where it needs to be (and I was under the impression that overcoming 'centrifugal force' was the only requirement for oil pressure really.

My Dad has done a calculation, taking a small mass of oil at several different radii, and adding up the pressures to come to a value of 36psi at 7000rpm. This adding up of chunks is essentially what integration is, which is why I thought it might be possible to get a more accurate figure via mathematics... And I still hope to (as I don't fully trust his method yet).

Interesting i got told the v8 supercars over here run around 35psi oil pressure...

You could always just ring Toms (or whoever manufactured your pumps) and ask them. Or, perhaps another driver in your class with a similar setup? lol all those formula's would have broken my mind long ago

So the force on a slice of oil between r and r+dr is:
df=slice mass*r.w^2
= v.A.dr.r.w^2

This force should be compensated by pressure.
df=P(r+dr)*A-P(r)*A

(P(r+dr)-P(r))=v.r.w^2.dr
dP=v.r.w^2.dr

So if P0 is the oil pressure you want at the center
P(R)=P0+(v.R^2.w^2)/2

With R=0.0275, v=810, w=523
P(R)=P0+12.15 psi

This result assumes the hole is filled all the time with new oil, steady state which is not the case in your drawing. There should be a large safety multiplier. But it seems pressure has to vary with square of angular speed and square of radius.

Thanks Juls, that was the sort of maths based analysis I was after, and it confirms that the method isn't really appropriate, as I feel that 12psi is far too low at 5000rpm (523rad/sec).

I think I'll probably aim for 30psi as a minimum at that speed, and just hope. It's not easy getting in touch with the racing department of Tom's, and I don't trust what other people do in the paddock as they don't know what they're doing themselves really, so it would be the blind leading the partially sighted.

I'll see what happens at Mallory in 10 days time. Would really like the get the win, but early days in the car at the moment, so who knows.

P.S. Anyone want to make me an avatar of my current car with the comic green cloak? I like it, and I want to keep it.

Euh just to unroll the thread.
I wonder if we can find the multiplier required if we take into account the fact that the hole receives oil only when it is in front of the oil entrance (at least partially).

Do you have an estimation of the diameter d of the hole? To chek if what I get can be useful.

i think the current estimates are off by enough that looking into other easons why oil pressure is necessary at all would be a better idea then to hold on to an i think obviously flawed understandong of it
maybe the pressure the piston exerts on the bearing during the work stroke which has to be cushioned by an equal back pressure from the oil

any easy way to get a good guestimate of the peak force that the piston gets pushed down with?

Quote from Shotglass :

i think the current estimates are off by enough that looking into other easons why oil pressure is necessary at all would be a better idea

Actually showing that was the intention behind the wall of text in my last post. The method shown there still needs a security factor of 5 to 10 and is already complex rather complex.
Really, I think any proper analytical solution, at least based on the given problem, should be filed under "ad absurdum".

Vain

IMHO, being as he has a ballpark to work off from the standard motor (albeit wet sump was something like 30-76psi wasn't it?) oil pressure @ 5000rpm, and u know the relief valve cuts in at 80psi, i'd say work out something in the middle of that you're happy with (so 30-40psi sounds pretty bang on, but i'm not a mechanic, and would take no responsibility for any info used lol. It does kinda make common sense, but getting hold of Tom's would be a big help, being as they would have a pretty good idea of what their dry sump kit should be doing on what motor at what revs (you'd think?)

If the drawing is correct (I mean it is not simplified and really works like that), and we consider oil is pushed only when openings are (at least partially) in front of each other.
Then the pressure received in the shaft is not anymore P, but (with some approximations, d small compared to D) P*2*d/(Pi*D).

So oil pressure required to obtain P0 in the center is now:
P=(P0+(v.R^2.w^2)/2)*(pi/2)*(D/d)

Did you ever get resolution on this tristan? Have you found a happy setup now?

No, I decided that the quantative analysis wasn't providing answers 'real' enough to be relied upon, so I'm going to use a combination of gut-feeling and the numbers from the manual - anything less than about 30psi at 5000rpm is too little...

The current problem seems to be a dodgy oil pressure sensor, which as of yesterday reads either zero or 21psi no matter what the revs are. But I think that's just a calibration error introduced during a recent dash firmware update...

The new clutch is in and bled, and although the pedal is now a smidgen lighter the available travel is a lot less, which might make slow speed driving in the paddock, as well as the starts, quite tricky. Easily fixed by a smaller master cylinder, but no time between now and the race...

An ex-F1 aerodynamicist is coming round in about 10 minutes to look at the car and make some suggestions for future upgrades regarding drag and downforce, and I think that'll be very interesting.

We've fiddled with the carbs a lot recently, as 2.0g is more than they seem to be able to stand before they cease controlling the mixture. Sadly, I'm regularly pulling 2.5g already, so it coughs and splutters out of corners a bit (felt more than heard, but if you watch onboard videos carefully you can see it happening). We think changed to floats, needle valves and air-corrector jets should have helped. Next year we hope to change to injection, and that will cure it instantly, but we're not allowed injection this year

New brake discs fitted, but more for peace of mind following the discovery of large (and worsening) cracks on the friction faces. I don't think it'll assist braking, and my current record of 1.9g is more than enough for the time being!!!

I'm not too confident going into Mallory, as it's a track I just don't gel with. Gerrard's is a long, long corner that is going to be damn close to flat out this year (scary stuff - but I did win the trophy for being the driver with "The Biggest Balls', so if anyone can do it flat it'll be me!), and the only other corner worth a damn is Shaw's hairpin, the slowest corner in UK motorsport, where the revs end up so low that drive out of the corner isn't great. But it's a bumpy track, with off-camber bits that seems to rewards a driving style I don't appear to possess. I'll have to do a Button and start flinging it around some more

Quote from tristancliffe :

I'm not too confident going into Mallory, as it's a track I just don't gel with. Gerrard's is a long, long corner that is going to be damn close to flat out this year (scary stuff - but I did win the trophy for being the driver with "The Biggest Balls', so if anyone can do it flat it'll be me!), and the only other corner worth a damn is Shaw's hairpin, the slowest corner in UK motorsport, where the revs end up so low that drive out of the corner isn't great. But it's a bumpy track, with off-camber bits that seems to rewards a driving style I don't appear to possess. I'll have to do a Button and start flinging it around some more

I think Mallory is a great little track, out of the three I've tried (Donnington and Silverstone GP), which are all tracks I'd rate highly from the other side of the fence, Mallory was right up there as far as I'm concerned. Gerrard's is a real challenge, got to keep your speed low enough to keep hugging the inside of the track, I found accelrating early was very tempting, when you've done the first 90 degrees you're not even half way round it, so throwing the car about really doesn't wok, you've got to set it up nicely and keep the forces sustained for a long time. Getting Gerrard's right is fantastic though, one huge four wheel drift in the Morgan, probably a totally different corner in your car though given that you're not going to be traction limited round the last half of it I guess you do the tighter bit as a corner in itself and the rest of it is just an undramatic exit onto a straight. The Esses is a nice section and definitely best to attack as one big block with smooth transition rather than trying to throw it from corner to corner, if you don't get a good exit from them you're bound to get passed on the run up to the hairpin.

The hairpin itself is horrible, it is quite literally like going round a mini-roundabout in the middle of a race track, too tight for just about any car to take otherwise, far too many cars seem to have problems getting round it getting carried away and spinning on the exit or understeering wide, invariably it's too fast an entry and overeggerness to put the power down that cause people these issues, whilst the car strolling round it in second gear at 30mph nips up the inside and blasts into the distance. Devil's Elbow is awesome, not quite flat in the Morgan, got to aim for the apex which is in the middle of the blind dip, feels like you're throwing yourself towards the inside wall, people trip up by not turning in properly and by getting out of shape panicing about the fact they're going to run wide, you can afford to run wide round Devil's Elbow, there's plenty of track that you need to use and if you run slightly wide off the track on the exit chances are you'll get away with it or just brush the wall. Get it wrong earlier in the corner and unfortunately it is one of the truely dangerous corners left in the country with very little run off and if you're not lucky enough to dig into the gravel trap your chances of walking away are low, certainly not a place for overtaking.

Quote from tristancliffe :

The current problem seems to be a dodgy oil pressure sensor, which as of yesterday reads either zero or 21psi no matter what the revs are. But I think that's just a calibration error introduced during a recent dash firmware update...

you should be able to test the voltage of that sensor, to eliminate it from the equation. I think there's specified voltages in the toyota manual, but i think you'd also need a manual oil pressure gauge to make sure actual pressure @ specified voltage is being achieved.

EDIT - i have asked the wise 3s gurus @ toymods for more info on this sender