BobbyZ

Well I think I am going to go ahead and replace the gaskets. It might be something else but this is the only idea I have. Does anyone have any suggestions for other stuff I should do while I have the manifolds off? Other gaskets to replace?

Hi all, I am still trying to get my 280zx turbo running well. When I first start the car (when it is cold) up it surges and bucks a little bit and then will start to run a little better. If I hit the gas, then the revs drop a little and it starts to stall. When it hits boost like this it makes a very loud whistling noise. Eventually when the engine is really hot, the sound will go away. All of this seems to point to a leak in the intake somewhere. Listening around the engine bay at idle it just started another noise which sounds like a loud vibration (maybe air moving past something that is rattling). This sound seems to be coming from the intake manifold near the head. My conclusion is that the intake manifold gasket has a leak somewhere, maybe the P.O. didn't tighten up the bolts enough. Once the head gets hot it will expand and seal against the intake manifold. Does this make sense to anyone else? My plan is to remove the intake manifold to replace the gasket tomorrow. If I do remove the intake manifold, is there anything else I should look at while I am in there?

Maybe a little my fault... Truce. Until next time...

I guess technically it does but a voltage meter has an impedance which is virtually infinite meaning there is still no current running in the circuit. In order for a resistor to do ANYTHING it has to have current running through it. That is incorrect, if there is current running through the circuit then the LED will "see" whatever voltage is left after the voltage drop across the resistor. This voltage drop across the resistor will change depending on the current. Here is my proof... Voltage at the supply is 5V Voltage at the LED is around 2.8V (Apparently that is the turn on voltage of these LED's, I just had them sitting on my desk so I dont know the specs.) Lets try the old analogy of water verses electricity. The analogous component in water flow is any kind of restriction. If you have water running through a hose and you put a kink in it, it will act like a resistor. The water pressure (voltage) will be lower after the kink than it was before the kink, the water flow (current) will also be lower in the entire hose. However this requires water flow. If you have the same hose with a kink in it but there is a nozzle that is shut at the end so no water is flowing, than the kink wont do anything. It wont reduce the water flow (already zero), and it wont reduce the water pressure. That is what your first circuit showed; no current running means no voltage drop across the resistor. You are completely correct in your evaluation of the circuits you posted most recently. The voltage drop across each resistor is going to be the same becuase you have scaled up your voltage divider (did you hear that "voltage drop across each resistor"). However, this does not mean that they are not dropping voltage. Your statement impede voltage does not make a whole lot of sense. Impede implies some kind of flow. Voltage is not a flow, it is a "pressure" in one sense. (How big of a difference in the number of electrons is there between two points) That is essentially what voltage means. You may not find a single textbook that says "resistors impede voltage". But every single textbook will say that voltage will drop across a resistor. (Specifically they will say that the V=IR) OHMS LAW!!!! Yes a resistor will limit current I agree with that statement, but your other statement that resistors do not drop voltage is incorrect. They drop voltage very well.

Absolutely beautiful. Your updates never dissapoint Monzter.

You have sponsorship!

The dude just ran off to catch the bus? Didnt even turn off the car first.

Technically I don't think that your first picture is a "circuit" the definition that I know requires a path for current. S: (n) circuit, electrical circuit, electric circuit (an electrical device that provides a path for electrical current to flow) The first picture you have has no current running so it is not dropping any voltage. The resistors aren't actually dropping any current either in your example. Yes resistors limit current (when there is current), but they also drop voltage. The LED example is perfect. Once an LED has turned on (forward voltage > 0.7V) I will step through how to choose a resistor for an LED. You need three things: Input Voltage, Voltage Drop of LED, and Current Required. The circuit is shown below. (Nasty schematic drawn in paint on a touchpad, I am on my wife's laptop so I dont have my fancy schematic capture software) E.G. (These are fairly typical) Vin = 5V Vf = 0.7V I = 10mA Voltage Drop across resistor = 5-0.7 = 4.3V Resistance required for current of 10mA R=V/I (Ohms Law Again!!) R=4.3/10mA = 430 Ohms You have to find that voltage drop across the resistor before you can find the resistor to use. There you have it check it here: http://led.linear1.org/1led.wiz if you dont believe me. (There answer is different because they round to the nearest real resistor value that you can buy) I just designed a PCB which required me to drive around 40 LED's and I had to choose the resistors carefully so that the LED's would all turn on and I wouldn't have to mess with my input voltage too much. (0.7V is just an example turn on voltage. The LED's I used on the board above actually had a voltage of 1.1V) Dont forget Ohm's Law...

Sorry I may have gone a little overboard with the pictures. Any chance to play with the Keithley sitting on my desk. That thing is so cool

I have no idea what that picture is showing. It looks like you are applying 14.6 volts across the lead. And then 14.6 volts across the resistor. What the display is showing is just the voltage you are applying. When you have the 14.6 volts across the resistor, one lead is at 14.6 volts and the other lead is at 0 volts. The resistor has dropped 14.6 volts at about 1.825 mA. Watch me! Its harder to see in my pictures but you get the idea. I am applying 5V to a series of 3 resistors current is 0.77mA. Watch the voltage drop (the little pointy things are the probes for the fluke meter) Voltage is about 5V (resistance in the wires has caused a voltage drop) After the first resistor the voltage has dropped about a volt. Then a large amount of voltage is then dropped through the second resistor That last resistor drops the final 0.62 volts, if I had probed the last point I would have gotten zero. Two can play at this game... Dont forget Ohms law. V = IR (The voltage drop across a resistor is equal to the current through it multiplied by the resistance) Resistors drop voltage just fine. I still think that the variable resistor could work fine. Like you said, the AFR works off of a voltage divider. You are just adding adjustment to the divider. Might be hard to get right but it seems like it could work.

Ya they do. The voltage on one end of the resistor will be lower than the voltage on the other. Current will also drop with the addition of a resistor. I think this idea might work fine. If you added a variable resistor to the line than the ECU is going to see less voltage and assume there is a smaller amount of air, adding a smaller amount of fuel. Tightening the AFM would correct for this and then you would have a variable resistor that increase or decrease the percieved amount of air that the ECU sees, changing the amount of fuel added. I think it would work fine, might be hard to find the correct range of resistance values to use without destroying the engine though.

First post says they will most likely (100%) require fender rolling or flares

Thats what it would look like if you jumped out of your car at 90mph

I drove the car to lunch and I am realizing that the compressor itself is making a lot of noise when the A/C is on. Once I turn the A/C off all I get is the noise from the A/C clutch. I may just pull the belt off again and get used to having the windows open. I think the only thing I could do is replace the compressor, not really an option due to the cost. (compressor + recharge = $$$)

I have been slowly fixing a few things on my newly aquired 280zx turbo whenever I get the time. This weekend I addressed the A/C. The previous owner had removed the belt to the A/C compressor at some point in this cars past and I wanted to see if it still worked. I put a new belt on there and the A/C seems to work great! However, when the A/C is turned off the clutch on the compressor makes a very loud rattle. It looks like the clutch is having a hard time freewheeling, you can see the compresser pully jerking around as the clutch catches and then lets go. The compressor may have sat for years without turning so I am assuming that the bearings are frozen on the clutch. When the A/C is engaged it is totally quiet so I am pretty sure that the compressor itself is fine. Can I spray the compressor clutch with lubricant to free it up or am I going to damage the clutch that way? Am I going to have to replace the clutch on the compressor?

I have heard great things about brown bread. I have heard better things about their newer stuff, the BXT.

Any insights on your experiences with it?

I thought it was faked until I watched it again... lol

I just did a google search and found some articles on the effects that the CF driveshaft had. http://www.stangpro.com/html/articles/driveshaft.htm One of the most interesting effects of the CF is the smoothness, apparently the carbon-fiber acts as a kind of damper for the vibrations from the diff. That would be kind of a cool way of making a DD a little more comfortable if it actually worked. If you actually do this make sure to give us an update, it seems expensive but I like the idea.

My bearings had the same problem. On mine the bearing was originally put in crooked and years of the suspension banging on it had jammed it in sideways. Eventually I am going to have to replace the entire assembly. For now I just left the bearing in there frozen, probably not a good idea

what is this "dumping" you speak of...

I am crying I am laughing so hard... at the office. Getting some funny looks and I cant really explain this one to anyone else.

Thanks for that grumpy, exact information I was looking for. I just wanted to see if you could do that without damaging the sensor.

I realize that this is not to be considered a replacement for a wideband. I am just looking for a cheap way to get a rough idea what the AFR's are in my engine. This would not be for tuning as much as troubleshooting. Here's a for instance, when my engine is warming up it has no power, ZERO I can barely move. I already have a narrowband installed in the exhaust of my car, and I also have a multimeter. I should be able to measure the voltage output of the O2 sensor to see if I am way rich because of a problem with the cold start circuit. I am looking at this as a replacement for buying a narrowband AFR meter.

Would it be possible to just use the stock narrowband sensor in the 280zx along with a multimeter. One of the sites listed above states that the output of a narrowband should be 0.5 volts at 14.7:1. I was thinking about adding some connections to the o2 sensor and wiring it into the cab of the car. That way when I make any changes to the engine (boost, porting, whatever) I can connect a multimeter to my test points inside the car and make sure that I am still getting enough fuel. Just looking for a cheap way to read out some AFR information, and this method is free. Eventually I could just build my own meter for the narrowband using LED's. But for now a multimeter seems like it would work just as well.