I am going to try several more times over the next few days just to be sure the vacuum method reliably works on a consistent basis. Unfortunately there are not a lot of really cold temperatures forecast.

As I got thinking logically through the details of how the vacuum might actually be assisting cold starts, I pulled a bottle of whisky out of the freezer where I store it and noted how much thicker it poured than at room temperature. I surmised that the low pressure created by the shop vac would have a chance to reach the carb at that brief moment of valve overlap at the top of the exhaust stroke. The vacuum would then draw fuel, which has thickened slightly from the cold, just like the whisky, through the jet orfaces.

The phenomenon of gasoline thickening at cold temperatures, just like the whisky, may be the root cause of the 'extreme' cold starting hesitation which the vacuum overcomes(?)...

 

Gasoline thickens? Novel thought.
I'll go with lack of vaccuum.
12V Dustbuster will be your friend, except that means an extra draw on the battery.

 

"Gasoline thickens? Novel thought.
I'll go with lack of vaccuum."

If it were purely lack of vacuum, I'd think the problem would persist at warmer temperatures too.

It's well understood that diesel gels up in extreme cold (hence the use of tank heaters up north). And certainly my Scotch is a lot thicker when it's stored in the freezer.

That starter jet is pretty small, even when drilled to .019" like mine is. Wouldn't take much change in viscosity befor the fuel wouldn't flow well though it. It's got me thinking if I shouldn't try drilling out the starter even a little more(?)...

 

I'll give IDRIDR & TNC first right of refusal before I head down to the patent office myself!

 

Is the starter jet always in the circuit? If it is you could be messing up your running mixture.

 

My understanding is that the starter circuit ONLY activates while the choke is engaged, otherwise it's dormant...

 

That is when you need to go to some sort of heat to warm the engine. Aim a spot light at the engine to warm it or figure out how to get some sort of block warmer on it. Happens with diesels, why wouldn't it happen with a regular four stroke single. I doubt my 650 would start at 10 degrees. It's all about the drag of the cold oil not allowing it to spin fast enough. If you didn't have the KACR to at least relieve some of the compression I guarantee it wouldn't start at as low a temp it does now. It would not only be fighting the oil drag, but it would have to go up against the full compression.

One other possibility is to go to a thinner oil, 5w30 or lower if you're going to ride at those temps.

 

My bet is cold starting doesn't have much of any heat generated by the compression in the first few hundred revolutions to make any difference. Taking off the KACR would just strain the battery that much more trying to pull the piston over TDC. It ain't a diesel and even if it was it would be requiring some sort of ignition system to aid the compression or have a block heater to get enough heat to start. There's a reason there are plugs hanging out the front of most school busses up in the north during the winter months...

 

There you go... put a block heater in it! Problem solved.

Diesel engines will fire without a plug once they get running due to compression. I believe they have compressions in the range of 16:1 to 20:1.
I submitted this question to the U of IL physics department, as I cannot believe my calculation.

 

OK, I got a response back from the U of IL physics department. The Q (question) and A (answer) are at the bottom of this post or in the next post by me.

We will use the calculation that they made using the 1:12 compression, as that is more of a hard number than the 135psi compression test info. Remember that these numbers are theoretical and actual numbers will vary, but not by more than a factor of .5 I would imagine.

Based on this the compression temperature if full compression would be about 1400degF.
I revised the numbers to reflect starting at 32deg F (273Kevin).... 273*2.7=737C=1359F

Using a reduced compression pressure of 80psi due to the KACR & assuming no loss in air (this is not actually true due to the exhaust valve lifting and letting air out of the cylinder) the temperature would be about 840degF.
273*1.64=448C=837F 1.64=80/14 'TO THE POWER OF' .285

THEREFORE IN CONCLUSION: It appears that the KACR is in fact lowering the chamber temperature quite a bit on the initial rotation. These temperatures would be less due to some of the heat being transmitted to the cylinder, head, piston, & final vaporization of fuel.

THOUGHTS: I think the KACR is limiting cold starting a little, but once started the engine is very cold blooded, which again points to the fact that there is not enough fuel in the air mixture at these temperatures. If you want to start at low temps, you will need to change the starting jet considerably or go to a pumper carb where you can add fuel by twisting the throttle.

Thanks & good night.

Q: I am trying to calculate the increase in temperature in a combustion engine due to compression. 250cc volume, 12 to 1 compression, initial pressure 14#/sqin, compression pressure 135#/sqin. Beginning temperature 30 degrees. Using ideal gas laws I come up with 4300 degrees, that cannot be right. Air is not ideal gas, but it should not be that far off.
- darrel kruger (age 62)
cottonwood, AZ, USA

A: The numbers you give aren't quite consistent. This is close to an adiabatic compression, meaning that it happens too rapidly for there to be much heat flow out of the gas. For an ideal gas with a temperature-independent heat capacity, we have that pV 'TO THE POWER OF' (γ) stays constant, where p is pressure and V is volume. Those conditions are pretty well obeyed by air, with γ =1.4. So for a factor of 12 compression you expect p to go up a factor of 12 'TO THE POWER OF' (1.4)=32.4. Yet you seem to say that p only goes up by 135/14= 9.6.

At any rate, for an ideal gas the absolute temperature T is proportional to pV, or in our ideal case V 'TO THE POWER OF' (1- γ) or p 'TO THE POWER OF' (γ-1)/γ. If we use your x12 compression in V, we'd get a factor of 2.7 for the increase in T. If we use the x9.6 in p, we'd get a factor of 1.9 increase in T.

When you say"Beginning temperature 30 degrees" I assume that's in °C, or T=303K. So you end up with T of 2.7 *303 K or 1.9*303K, depending on whether your V or p numbers for compression were right. You can do the calculations more carefully, but that's in the range of about 300°C to 540°C.

Mike W.