Why bigger turbos make more HP at the same PSI....
04-23-07, 04:53 PM
#1
Why bigger turbos make more HP at the same PSI....
My explanation: Not to be taken as gospel ;o)
I struggled to wrap my mind around this for a while.
example thought process: You have an engine which rotates at a given RPM, let's say 6000 rotations per minute for grins. now at 6000 rpm the intake port is open for the same amount of time for both a t78 and a stock twin setup. So the time intervals are fixed by the engines rotation. Assuming that both setups are creating 10 PSI of pressure measures from the intake manifold (which has a fixed internal volume) then it seems impossible for one turbo to make more power at 10 PSI than another because both the volume of the engine and the intake manifold are fixed. And since the volumes are fixed that means the same net force is exerted onto the intake charge in both scenarios.
this is the point where we scratch our heads ^^^
Most of you know this but I'll say it anyway:
1PSI = 1LB force per square inch (not Pounds of air per square inch!) - think about it, a square inch is a unit of area, not volume. 10 PSI = ten pound of force exerted exerted on every square inch of internal surface area of the intake manifold and intake ports = says nothing about how much air is in the intake/engine ( if it did it would be per cubic inch) just how much force the air is exerting as it gets force fed from the turbo's compressor.
With that being said, one can calculate the air density based on how much pressure is exerted, but PSI is not a measure of volume in and of itself.
Now, "Cubic feet per minute (CFM) is a non-SI unit of measurement of gas-flow (most often air-flow) that indicates how many cubic feet of gas (most often air) pass by a stationary point in one minute. In other words, it is a unit for measuring the rate of flow of a gas or air volume into or out of a space."
-wikipedia
OK, so we accept that a large turbo expells air at a higher velocity and therefore has a higher CFM. Now lets try to understand why:
One easy way to understand how larger turbos move air at a higher velocity it is useful to think about a neck in a river and the way that water flow through it.
The wheel size and outlet volume of a T78 turbo compressor is much larger than that of the twin setup. This large volume of air leaves the turbo and enters the bottleneck which is the intake tract speeding up just as water speeds up in a river bottleneck.
The air flowing from the small twin compressors on the other hand are flowing into the same river, but this time the river is large in relation to the charge volume so the air just creeps along.
OK so we established the following:
The tricky part:
At 10PSI (assuming the same intake temps) the T78 is cramming the same amount of air into the intake manifold as the twin setup. So what gives?
Here's the key point to understand:
The boost that you are reading at the intake manifold is not telling you how much air actually makes it into the intake ports during their short open interval. It's only a measurement of force exerted on the intake plenum.
Key to understanding:
The air coming from the twins will surge forward into the intake ports with a lower velocity than that from the T78 for the reasons that we established in our "river bottlnecK" example earlier.
So while both turbos are exerting the same amount of force on the intake ports the air from the T78 is approaching the intake ports at a higher velocity and therefore more will get in before the port closes.
Lastly: The boost reading that you see on your gauge is not taken from within the engine, it is taken from your intake plenum and should not be confused with an internal measurement. Understand these points and you will have the problem cracked.
I may be wrong but that's my logic ;o) Hope that helps someone.
I struggled to wrap my mind around this for a while.
example thought process: You have an engine which rotates at a given RPM, let's say 6000 rotations per minute for grins. now at 6000 rpm the intake port is open for the same amount of time for both a t78 and a stock twin setup. So the time intervals are fixed by the engines rotation. Assuming that both setups are creating 10 PSI of pressure measures from the intake manifold (which has a fixed internal volume) then it seems impossible for one turbo to make more power at 10 PSI than another because both the volume of the engine and the intake manifold are fixed. And since the volumes are fixed that means the same net force is exerted onto the intake charge in both scenarios.
this is the point where we scratch our heads ^^^
Most of you know this but I'll say it anyway:
1PSI = 1LB force per square inch (not Pounds of air per square inch!) - think about it, a square inch is a unit of area, not volume. 10 PSI = ten pound of force exerted exerted on every square inch of internal surface area of the intake manifold and intake ports = says nothing about how much air is in the intake/engine ( if it did it would be per cubic inch) just how much force the air is exerting as it gets force fed from the turbo's compressor.
With that being said, one can calculate the air density based on how much pressure is exerted, but PSI is not a measure of volume in and of itself.
Now, "Cubic feet per minute (CFM) is a non-SI unit of measurement of gas-flow (most often air-flow) that indicates how many cubic feet of gas (most often air) pass by a stationary point in one minute. In other words, it is a unit for measuring the rate of flow of a gas or air volume into or out of a space."
-wikipedia
OK, so we accept that a large turbo expells air at a higher velocity and therefore has a higher CFM. Now lets try to understand why:
One easy way to understand how larger turbos move air at a higher velocity it is useful to think about a neck in a river and the way that water flow through it.
The wheel size and outlet volume of a T78 turbo compressor is much larger than that of the twin setup. This large volume of air leaves the turbo and enters the bottleneck which is the intake tract speeding up just as water speeds up in a river bottleneck.
The air flowing from the small twin compressors on the other hand are flowing into the same river, but this time the river is large in relation to the charge volume so the air just creeps along.
OK so we established the following:
- At a given RPM the intake ports are open for a defined period of time.
- A large turbo moves air more rapidly through an intake tract with a fixed volume.
The tricky part:
At 10PSI (assuming the same intake temps) the T78 is cramming the same amount of air into the intake manifold as the twin setup. So what gives?
Here's the key point to understand:
The boost that you are reading at the intake manifold is not telling you how much air actually makes it into the intake ports during their short open interval. It's only a measurement of force exerted on the intake plenum.
Key to understanding:
The air coming from the twins will surge forward into the intake ports with a lower velocity than that from the T78 for the reasons that we established in our "river bottlnecK" example earlier.
So while both turbos are exerting the same amount of force on the intake ports the air from the T78 is approaching the intake ports at a higher velocity and therefore more will get in before the port closes.
Lastly: The boost reading that you see on your gauge is not taken from within the engine, it is taken from your intake plenum and should not be confused with an internal measurement. Understand these points and you will have the problem cracked.
I may be wrong but that's my logic ;o) Hope that helps someone.
04-23-07, 07:58 PM
#4
Rob, I'm sure CarbonR1 can correct me, but basically in a nutshell, PSI is pressure. Pressure is defined as force over surface area, correct?
To oversimplify, while the pressure (psi) may be the same when comparing two different turbos, the surface area (air the turbo can hold and thus blow) is completely dependent on the size of the wheel, housing etc (for simplicity's sake we'll just say dependent upon the size of the turbo, period).
So at a given pressure, a large and small turbo will move different amounts of air b/c their surface area is different. And if one has a much larger surface area, the resultant force from the turbo is greater, which leads to greater forced induction of air into the IC --> more cold air --> greater density of O2 into the motor --> greater combustion --> greater power.
If you look at it from the opposite end, a small turbo can be moving a very small amount of air (due to it's small surface area) and a larger turbo can be moving a HUGE amount of air (due to it's large surface area). As such, the power levels produced by each turbo will be different, despite the pressure being the same. The reason being is that b/c the small turbo has a small area, it takes LESS air to reach a higher pressure rating. A larger turbo requires much MORE air to reach that pressure rating. So indeed the volume of air moved is quite different, even at the same pressure rating (PSI).
One more.
I can have a small balloon w/ a LOT of air packed into it. The pressure is very high, given that even though there is LITTLE air in the balloon, b/c the surface area of the balloon itself is small, the molecules in the balloon don't have much room to move around, so they bump into one another and into the walls of the balloon more often, resulting in a very high PRESSURE.
Likewise, I can have a GIGANTIC airpump, which has a HUGE amount of air, MUCH more than the balloon, but it has the same pressure. In fact, it would take a LOT more air to reach that higher pressure simply b/c the airpump's surface area is so huge. In fact, you can have the airpump at a lower pressure than the balloon, and the airpump still has MUCH more air in it than the balloon - but those molecules just aren't as tightly packed, simply b/c they have more room to move/play
So in short, the two turbos are NOT cramming the same AMOUNT of air into the engine simply b/c they're running at the same PSI. The amount of air (which they're pushing into the engine) is just pressurized to the same degree, although the AMOUNT itself may (and in your example indeed does) differ.
Kinda a different example. I can apply the same exact force by poking you. But if I poke you w/ my fist, I probably won't hurt you. But if I poke you holding a thin needle, I'm gonna hurt ya much. Why? B/c it's dependent upon the surface area the force is being applied to, which translates to the PRESSURE. The pressure exerted by the needle is FAR greater than the pressure exerted by the fist. Yet the force may be identical.
Does that help any (in layman's terms)?
~Ramy
To oversimplify, while the pressure (psi) may be the same when comparing two different turbos, the surface area (air the turbo can hold and thus blow) is completely dependent on the size of the wheel, housing etc (for simplicity's sake we'll just say dependent upon the size of the turbo, period).
So at a given pressure, a large and small turbo will move different amounts of air b/c their surface area is different. And if one has a much larger surface area, the resultant force from the turbo is greater, which leads to greater forced induction of air into the IC --> more cold air --> greater density of O2 into the motor --> greater combustion --> greater power.
If you look at it from the opposite end, a small turbo can be moving a very small amount of air (due to it's small surface area) and a larger turbo can be moving a HUGE amount of air (due to it's large surface area). As such, the power levels produced by each turbo will be different, despite the pressure being the same. The reason being is that b/c the small turbo has a small area, it takes LESS air to reach a higher pressure rating. A larger turbo requires much MORE air to reach that pressure rating. So indeed the volume of air moved is quite different, even at the same pressure rating (PSI).
One more.
I can have a small balloon w/ a LOT of air packed into it. The pressure is very high, given that even though there is LITTLE air in the balloon, b/c the surface area of the balloon itself is small, the molecules in the balloon don't have much room to move around, so they bump into one another and into the walls of the balloon more often, resulting in a very high PRESSURE.
Likewise, I can have a GIGANTIC airpump, which has a HUGE amount of air, MUCH more than the balloon, but it has the same pressure. In fact, it would take a LOT more air to reach that higher pressure simply b/c the airpump's surface area is so huge. In fact, you can have the airpump at a lower pressure than the balloon, and the airpump still has MUCH more air in it than the balloon - but those molecules just aren't as tightly packed, simply b/c they have more room to move/play
So in short, the two turbos are NOT cramming the same AMOUNT of air into the engine simply b/c they're running at the same PSI. The amount of air (which they're pushing into the engine) is just pressurized to the same degree, although the AMOUNT itself may (and in your example indeed does) differ.
Kinda a different example. I can apply the same exact force by poking you. But if I poke you w/ my fist, I probably won't hurt you. But if I poke you holding a thin needle, I'm gonna hurt ya much. Why? B/c it's dependent upon the surface area the force is being applied to, which translates to the PRESSURE. The pressure exerted by the needle is FAR greater than the pressure exerted by the fist. Yet the force may be identical.
Does that help any (in layman's terms)?
~Ramy
Last edited by FDNewbie; 04-23-07 at 08:04 PM.
04-23-07, 08:03 PM
#5
Basically just explained the efficiency differences between various turbos affecting intake temps intake temps and exhaust restriction to arrive at the same conclusion. Almost a prequel really. I think my velocity explanation is incomplete without taking exhaust backpressure into account (especially since my explanation does not account for different compressor RPMs) so I appreciate the link, especially the last post. Sometimes when you try to get too specific you forget to mention obvious. lol though I will admitt I never thought larger turbos to be less restrictive because of the harder-to-turn turbines and greater parasitic drag of a large compressor wheel.
04-23-07, 08:16 PM
#6
Originally Posted by FDNewbie
Rob, I'm sure CarbonR1 can correct me, but basically in a nutshell, PSI is pressure. Pressure is defined as force over surface area, correct?
To oversimplify, while the pressure (psi) may be the same when comparing two different turbos, the surface area (air the turbo can hold and thus blow) is completely dependent on the size of the wheel, housing etc (for simplicity's sake we'll just say dependent upon the size of the turbo, period).
So at a given pressure, a large and small turbo will move different amounts of air b/c their surface area is different. And if one has a much larger surface area, the resultant force from the turbo is greater, which leads to greater forced induction of air into the IC --> more cold air --> greater density of O2 into the motor --> greater combustion --> greater power.
To oversimplify, while the pressure (psi) may be the same when comparing two different turbos, the surface area (air the turbo can hold and thus blow) is completely dependent on the size of the wheel, housing etc (for simplicity's sake we'll just say dependent upon the size of the turbo, period).
So at a given pressure, a large and small turbo will move different amounts of air b/c their surface area is different. And if one has a much larger surface area, the resultant force from the turbo is greater, which leads to greater forced induction of air into the IC --> more cold air --> greater density of O2 into the motor --> greater combustion --> greater power.
If you look at it from the opposite end, a small turbo can be moving a very small amount of air (due to it's small surface area) and a larger turbo can be moving a HUGE amount of air (due to it's large surface area). As such, the power levels produced by each turbo will be different, despite the pressure being the same. The reason being is that b/c the small turbo has a small area, it takes LESS air to reach a higher pressure rating. A larger turbo requires much MORE air to reach that pressure rating. So indeed the volume of air moved is quite different, even at the same pressure rating (PSI).
One more.
I can have a small balloon w/ a LOT of air packed into it. The pressure is very high, given that even though there is LITTLE air in the balloon, b/c the surface area of the balloon itself is small, the molecules in the balloon don't have much room to move around, so they bump into one another and into the walls of the balloon more often, resulting in a very high PRESSURE.
Likewise, I can have a GIGANTIC airpump, which has a HUGE amount of air, MUCH more than the balloon, but it has the same pressure. In fact, it would take a LOT more air to reach that higher pressure simply b/c the airpump's surface area is so huge. In fact, you can have the airpump at a lower pressure than the balloon, and the airpump still has MUCH more air in it than the balloon - but those molecules just aren't as tightly packed, simply b/c they have more room to move/play
I can have a small balloon w/ a LOT of air packed into it. The pressure is very high, given that even though there is LITTLE air in the balloon, b/c the surface area of the balloon itself is small, the molecules in the balloon don't have much room to move around, so they bump into one another and into the walls of the balloon more often, resulting in a very high PRESSURE.
Likewise, I can have a GIGANTIC airpump, which has a HUGE amount of air, MUCH more than the balloon, but it has the same pressure. In fact, it would take a LOT more air to reach that higher pressure simply b/c the airpump's surface area is so huge. In fact, you can have the airpump at a lower pressure than the balloon, and the airpump still has MUCH more air in it than the balloon - but those molecules just aren't as tightly packed, simply b/c they have more room to move/play
So in short, the two turbos are NOT cramming the same AMOUNT of air into the engine simply b/c they're running at the same PSI. The amount of air (which they're pushing into the engine) is just pressurized to the same degree, although the AMOUNT itself may (and in your example indeed does) differ.
Kinda a different example. I can apply the same exact force by poking you. But if I poke you w/ my fist, I probably won't hurt you. But if I poke you holding a thin needle, I'm gonna hurt ya much. Why? B/c it's dependent upon the surface area the force is being applied to, which translates to the PRESSURE. The pressure exerted by the needle is FAR greater than the pressure exerted by the fist. Yet the force may be identical.
Does that help any (in layman's terms)?
~Ramy
Does that help any (in layman's terms)?
~Ramy
04-23-07, 08:39 PM
#7
Originally Posted by RacerXtreme7
Turbo effiency plays major roll which, if you been reading I and many stated earlier means cooler air charge which equals greater charge density. More over as stated several times now... turbine effiency. A turbo in its effiency range requires LESS energey to turn over/spin at that flow/pressure verses one that is out of effiency. There by less back pressure. Larger frame turbos by nature have lower back pressure from the get go also most aftermarket turbos (not all) use newer more effiencient wheels to go along with better designed manifolds (not always the case) and who the hell upgrades a turbo without an upgraded exhaust which plays HUGE roll in spool and peak power. Its several factors that make a tiny out of range turbo thrashing and heating air at 15 psi verses a large huffer at 78% effiecency with a large effiencient turbine and littler restricted exhaust make gobs more power then the stocker. If you notice and look at a lot of dyno results, the middle frame turbos make the same power as the larger ones at moderate boost levels (1 bar or 14.7 psi). Example would be 60-1,62-1, t61, gt35R, GT40r, T78, T66, T04R all make roughly 400 RWHP at 1 bar of boost. They all are roughly at 74~78% effieciency range (on a mild ported 13b). Its boost above 1 bar were the 60-1, 62-1 fall flat and the other larger turbo's begin to wake up and make some power. The smaller turbo's spool faster. This is why I don't understand those throwing BIG *** turbos on there rides when they only run pump gas and limit boost to 15 psi or so. There making boost later then smaller turbos and making same peak power. In otherswards, they'd loose the race (less power under the curve). I guess a lot of guys use big as turbos to make up for other short commings?? Maybe they like the fact that if they throw some race gas in the tank and turn the wick up they know they'll have the huffer to back it (though this cenario rarely happens, and when it does its for a dyno plot to brag over)??
Sorry for the rant, its late and I have had more then 1 cocktail
~Mike...........
Sorry for the rant, its late and I have had more then 1 cocktail
~Mike...........
In a nutshell Mike explains how turbo size/efficiency affect backpressure (and charge temps which is obvious but worth mentioning ;oP) which in turn affect intake velocity.
Basically, disregard my river example and keep the rest (because it assumes matching RPMs - IOW only explains compressor efficiency, not velocity) and apply the backpressure/efficiency explanation to understand how velocity increases, and then realize that a turbo operating in its efficiency zone will generate less heat and therefore reduce charge temps to create a denser charge which equates to more power.
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04-23-07, 08:56 PM
#8
development
My nutshell reminds the student that there are about 7x 90 degree turns in the [opposing] stock twins intake path before the IC; and only Mazda and Obi-Wan Kenobi know WTF is going on between the exhaust manifold and turbine housings...then for the finally...A direct 90 degree turn right out the gate.
Exhibit A:
Exhibit A:
Last edited by dgeesaman; 04-24-07 at 07:00 AM. Reason: nobody misspells Obi-Wan Kenobi and gets away with it!!
04-23-07, 09:00 PM
#9
Originally Posted by dubulup
My nutshell reminds the student that there are about 7x 90 degree turns in the [opposing] stock twins intake path before the IC; and only Mazda and O.B. One Kanobie know WTF is going on between the exhaust manifold and turbine housings...then for the finally...A direct 90 degree turn right out the gate.
Don't speak to loudly, the OEM mongers will hear you.
04-24-07, 12:12 AM
#13
2/4 wheel cornering fiend
Originally Posted by 1.3 liter V8 eater
in a nutshell... or in a rats nest?
Originally Posted by Monkman33
does adiabatic efficiency come into play anywhere?
04-24-07, 10:33 AM
#14
Originally Posted by wanklin
That assumes the same RPM
Irrespective of the rpm, the point still remains: same psi doesn't mean anything unless the turbos are the same SIZE. If the turbos are different sizes, then you can't compare psi at all. Apples to oranges, b/c the volume being moved is completely different.
It's the opposite. The larger turbo has to do less work to pressurize the intake charge, hence the greater efficiency. I think the link above was helpful in explaining that reduced exhaust restriction helps flow velocity from the back end, as well as the lower charge temps.
~Ramy
04-24-07, 02:44 PM
#18
Originally Posted by FDNewbie
RPM can be left out of the equation actually. The thing is, the larger the area of the turbo, the greater the air it hold/push, right?
So as such, it'll take either longer (timewise) or faster (rpmwise) (or both) to fill that area. So 14psi on a small turbo may take 2 - 3 seconds at 2500 rpm, while a large turbo may take say 5 - 6 seconds at 4500 rpm.
If the turbo has not spooled you will not get sufficient boost so the issue is null and void. I think this whole discussion is based on the assumption that the turbo has spooled up already in both scenarios. Accepting this, A small turbo rotating at a higher RPM can achieve the same flow as a larger turbo rotating at a lower RPM if the exhaust back pressure and intake tempts are held constant. RPM does play a role.
If they were rotating at the same speed the larger turbo would compress more air and higher manifold pressure would result.
Irrespective of the rpm, the point still remains: same psi doesn't mean anything unless the turbos are the same SIZE. If the turbos are different sizes, then you can't compare psi at all. Apples to oranges, b/c the volume being moved is completely different.
I wasn't talking about work or efficiency actually. But you're right that a larger turbo can work less and thus more efficiently to move the same amount of air a smaller turbo would. But regardless of the turbo, at some higher psi, the efficiency will drop and work will increase.
~Ramy
~Ramy
04-24-07, 02:56 PM
#20
Yes, his car resembles his posts, clean, simple and to the point. I like the color scheme as well BTW. Textured black is the way of the future AFAIC. Is that paint on your IC or powder dubulup? I decided to anodize my whole rad black so we'll see how that goes. lol
04-24-07, 03:31 PM
#23
On flats
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well let's simplify this. . .ALOT.
PV=nRT -> PV/nRT = 1 -> P1V1/n1R1T1 = P2V2/n2R2T2
For a given setup. . .ALL things except turbos identical, for the same boost
P1 = P2
V1 = V2
R1 = R2 = constant
therefore
n1T1 = n2T2 where n is the mass of air.
Let's call the smaller turbo 1 and the larger turbo 2
n2 = n1T1/T2
in other words. . .the mass of air that the larger turbo pushes at the same pressure, all things being equal, is equal to the mass of air the smaller turbo pushed times the ratio of temperatures. Basically, the adiabatic efficiency with which the larger turbo can compress n2 amount of air is greater such that the temperature of the compressed air is less so that the above relationship is satisfied.
Let's look at it another way, holding the same amount of air each is pushing constant (same power output assuming equivalent tuning since the mass of air is what dictates power made assuming equivalent combustion (tuning)).
P1V1/n1R1T1 = P2V2/n2R2T2
simplifies to
P1/T1 = P2/T2
P2 = P1T2/T1
again, since the larger turbo can compress the same mass of air more easily, less work is done, less heat is generated, T2 is less, therefore the ratio T2/T1 is less than one, therefor P2 < P1, therefore, the larger turbo makes more power (pushes a larger MASS of air) at a lower pressure than the smaller turbo.
This is a really simplistic way to look at it, but most of the assumptions of the ideal gas law are valid *enough* for this to work. there are much more complex dynamics at work that both compliment and counteract, but for basic understanding's sake. . .it'll work.
Now everyone bust out your physics 1 book and study up.
Or I'm just wrong and a hugely pompous ***. . .either way
ryan
PV=nRT -> PV/nRT = 1 -> P1V1/n1R1T1 = P2V2/n2R2T2
For a given setup. . .ALL things except turbos identical, for the same boost
P1 = P2
V1 = V2
R1 = R2 = constant
therefore
n1T1 = n2T2 where n is the mass of air.
Let's call the smaller turbo 1 and the larger turbo 2
n2 = n1T1/T2
in other words. . .the mass of air that the larger turbo pushes at the same pressure, all things being equal, is equal to the mass of air the smaller turbo pushed times the ratio of temperatures. Basically, the adiabatic efficiency with which the larger turbo can compress n2 amount of air is greater such that the temperature of the compressed air is less so that the above relationship is satisfied.
Let's look at it another way, holding the same amount of air each is pushing constant (same power output assuming equivalent tuning since the mass of air is what dictates power made assuming equivalent combustion (tuning)).
P1V1/n1R1T1 = P2V2/n2R2T2
simplifies to
P1/T1 = P2/T2
P2 = P1T2/T1
again, since the larger turbo can compress the same mass of air more easily, less work is done, less heat is generated, T2 is less, therefore the ratio T2/T1 is less than one, therefor P2 < P1, therefore, the larger turbo makes more power (pushes a larger MASS of air) at a lower pressure than the smaller turbo.
This is a really simplistic way to look at it, but most of the assumptions of the ideal gas law are valid *enough* for this to work. there are much more complex dynamics at work that both compliment and counteract, but for basic understanding's sake. . .it'll work.
Now everyone bust out your physics 1 book and study up.
Or I'm just wrong and a hugely pompous ***. . .either way
ryan
04-24-07, 03:54 PM
#24
development
Originally Posted by rynberg
I've always admired how efficiently you ran the piping and intake on your setup.
Originally Posted by wanklin
Yes, his car resembles his posts, clean, simple and to the point. I like the color scheme as well BTW. Textured black is the way of the future AFAIC. Is that paint on your IC or powder dubulup? I decided to anodize my whole rad black so we'll see how that goes. lol
Powder coated Wrinkle black is the cleanest thing under a hood...IC is paint. I always wondered, which was better for keeping heat out (or letting heat out??) powder or anodize
To: Big_Rizzlah
the way I think of it is...big turbo pushes air easily and efficiently thru intake, and exit easily thru exhaust = less backpressure on engine...rinse and repeat complete circle.