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The stock air box's snorkel intake is not really restrictive. However if one makes the assumption that is and drills holes to somehow try to let more c.f.m. in ( it really won't), why would one think this extra cubic feet per minute of air will somehow make the bike run leaner?
If one assumes one gets higher air flow rates through the carburetor it will just stimulate more fuel to be sucked through various fuel circuits resulting in same stoichiometric ratios, just at a higher c.f.m. One has to remember the Bernoulli principle and how it makes our carburetors work. Increasing the air flow rate for any given throttle opening will be accompanied by a corresponding increase in the fuel flow rate; it will not alter the air/fuel ratio which is determined by all the fuel passageway orifice sizes, not air flow rates.
I know there is lots of shade tree wisdom and old wives tales repeated out there, also lots of gimmicks for sale. Doesn't mean they work.
If one wants to lean out the air/fuel mix one needs to change the fuel metering to reduce fuel delivery rates. Simply trying to cram more air through simply grabs more fuel with that additional air; the ratio is mechanically fixed by the physical geometry of the various fuel orifices all driven by suction caused by the air flow according to the Bernoulli principle. Besides the drilled holes let more induction noise out while allowing a tiny bit more debris and air in.
 

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The stock air box's snorkel intake is not really restrictive. However if one makes the assumption that is and drills holes to somehow try to let more c.f.m. in ( it really won't), why would one think this extra cubic feet per minute of air will somehow make the bike run leaner?
If one assumes one gets higher air flow rates through the carburetor it will just stimulate more fuel to be sucked through various fuel circuits resulting in same stochiometric ratios, just at a higher c.f.m. One has to remember the Bernoulli principle and how it makes our carburetors work. Increasing the air flow rate for any given throttle opening will be accompanied by a corresponding increase in the fuel flow rate; it will not alter the air/fuel ratio which is determined by all the fuel passageway orifice sizes, not air flow rates.
I know there is lots of shade tree wisdom and old wives tales repeated out there, also lots of gimmicks for sale. Doesn't mean they work.
If one wants to lean out the air/fuel mix one needs to change the fuel metering to reduce fuel delivery rates. Simply trying to cram more air through simply grabs more fuel with that additional air; the ratio is mechanically fixed by the physical geometry of the various fuel orifices all driven by suction caused by the air flow according to the Bernoulli principle. Besides the drilled holes let more induction noise out while allowing a tiny bit more debris and air volume in.
Anyone's seat-of-the-pants opinion seems more wishful thinking than anything that can be substantiated by imperial evidence and the physics of fluid mechanics.
Fred,

Despite your thinking that drill holes in the airbox cover won't work, I'm telling you it does. As it was about 10 years ago when I tried it, I can't for the life of me remember who it was here on the forum that suggested this idea with the airbox holes.

But I did indicate the best way was to re-jet and the airbox holes should be considered only temporary.

If you get a spare airbox cover, try drill holes in it and see if you can tell the difference. I'll even loan you my modified airbox for you to test. Now, I'm not gonna say you will fall off the back of your TW when you crack the throttle, but you will be able to hear the difference even at idle with a slightly higher rpm.
 

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I can not see the TW200 able to utilize a less restrictive air flow than what is stock. The motor just isn't capable of pumping any more air through than is already available. I recommend plugging the holes and maintaining the air filter regularly.
 

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take your air filter lid and filter out and and try to ride youe bike down the road wide open. It will prolly start spitting and spuddering over about half throttle. It will make a difference..
 

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I find myself in the curious position of both agreeing with Fred, and agreeing with Admiral — though Fred did temporarily make my brain hurt with “stimulate more fuel to be sucked through various fuel circuits resulting in same stochiometric ratios, just at a higher c.f.m” — although he did redeem himself with “the ratio is mechanically fixed by the physical geometry of the various fuel orifices all driven by suction caused by the air flow according to the Bernoulli principle” — that at least I could (barely) understand.

Fred — yes — your theory works in “principle”

And yet I can “see” how Admiral’s theory would work in practice — otherwise why would we need to re-jet when putting K&N type filters on there ?

If you have a finite maximum fuel flow (determined by the main jet size), then increase the air-flow (the holes) - or more realistically “reduce the resistance to air flow” — then that fuel flow still remains finite.

It may not apply until the main jet is “full on” — but you can see where my thinking is coming from ……

(That’ll teach the pair of you to give me a headache) …….
 

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Fred,

Despite your thinking that drill holes in the airbox cover won't work, I'm telling you it does. As it was about 10 years ago when I tried it, I can't for the life of me remember who it was here on the forum that suggested this idea with the airbox holes.

But I did indicate the best way was to re-jet and the airbox holes should be considered only temporary.

If you get a spare airbox cover, try drill holes in it and see if you can tell the difference. I'll even loan you my modified airbox for you to test. Now, I'm not gonna say you will fall off the back of your TW when you crack the throttle, but you will be able to hear the difference even at idle with a slightly higher rpm.
well said Kris..........and I even understood every word you used
 

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I wouldn't think that changes made to the intake would have a great effect on the performance as the extra air has no where to go, as the exhaust is so restrictive. it is like a metering device. I can see there would be some changes as Admiral was saying
 

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Sorry for the argument guys, I was just trying to correct terminology. Remember I was responding to the assertion in post #4 that increasing thevolume of air through the air box via holes creates the solution to high altitude woes.
I assumed concern was over a overly rich air/fuel mix due to reduced air density at higher elevations. Right, we all on the same page?
Thus was proposed my challenge for someone to explain how increased airflow volume through the air box results in a leaner air/fuel mix. No one has yet addressed this. Your comments instead have correctly addressed pressure issues, or minor changes in the head loss across the air filter/ air box assembly. Easy to interchange these concepts, I was just trying to clarify the principles so the discussion would be a bit more accurate. Had Admiral spoke of the holes reducing pressure losses for a given airflow I would have given a thumbs up.
 

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Helmet lock should have been OEM equipment keyed to match your ignition/gas cap key. Lock assembly should have been riveted to frame. If missing try looking on Amazon or motorcycle aftermarket vendors for an alternate helmet lock solution.
 

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Fred.. assuming your theories are correct ( and I may not have understood them) then why would a person need to reject a carb with a muffler change?
 

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Remember we are taking about carburetors which at their most simplest work on pressure differences across a venturi. No mass air sensors or O2 sensors as in a fuel injected engine. Anything that affects the pressure on the either immediate side of a venturi affects the relative pressure difference between the two sides, and thus the suction generated, and thus the flow rate of the fuel sucked through venturi into the intake air stream. Thus the pressure differentials can be affected by changes to either upstream, or downstream geometry. With the exhuast we are now just talking about the other side of the same coin.
 

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Here, let's get really confused:
[h=2]BOYLE'S LAW[/h]Boyle's Law is an expression of the relationship between the pressure and volume of a fixed quantity of gas. It was initially described by Robert Boyle in the 17th century.
Key Points:

  • Temperature and moles of gas are constant
  • Graph is hyperbolic (see below) and asymptotic to both axes
  • Pressure and volume are inversely proportional to each other




Problems that require you to use Boyle's Law will only mention pressure and volume. Do not get fooled if the phrase "constant temperature" is used. This means that temperature remains constant and is irrelevant to the mathematics of the problem. An example problem may read:
Find the pressure on 5.25 L of gas that was originally 3.12 L at 1.54 atm.

In this problem, a preliminary conversion of units will not be necessary unless the problem explicitly asks for a unit that is different than that given in the problem. Refer to the Boyle's law equation above. Set aside P1 and V1 for the initial conditions of the gas. V2 will be for the new volume of the gas and P2 will be what we solve for:


[h=2]CHARLES'S LAW[/h]Key Points:

  • Pressure and moles of gas are constant
  • Graph is linear (see below)
  • Volume and temperature are directly proportional to each other




It's important to remember that temperature must be converted to kelvin when utilizing any of the gas laws. Since thermometers are designed to use degrees Celsius, lab data or values given in a problem will likely need a conversion. Note that in the example problem below, nothing is said about pressure:
A 5.0 L vessel of gas is held at 25°C. What will be the new volume if the temperature is doubled?
Do not be fooled into thinking that since the temperature doubles, so does the volume. That would be true if the kelvin temperature doubled, but the Celsius temperature doubling from 25 to 50°C is not a significant increase:



[h=2]GAY-LUSSAC'S LAW[/h]Key Points:

  • Volume and moles of gas are constant
  • Graph is linear (see below)
  • Pressure and temperature are directly proportional to each other




Consider the following problem as an example:
25.0 L of a gas is held in a fixed container at 1.25 atm at 20°C. What will be the pressure of the gas if the temperature is increased to 35°C?
First, the temperatures must be converted to kelvin:



Next, the appropriate subsitutions can be made into the equation. Note that the volume given in the problem is immaterial to the solution. The phrase "fixed volume" communicates that volume is constant in this problem, and thus any equation that uses "V" is not to be used.


[h=2]COMBINED GAS LAW[/h]

The combined gas law integrates Boyle, Charles, and Gay-Lussac's laws. Here, the only constant is the number of moles of gas. Notice that if you cover on set of variables, either Charles, Boyle, or Gay-Lussac's Law remains. For example, if you cover T1 and T2, the remaining equation is the same as Boyle's Law. Removing P1 and P2 leaves Charles's Law and eliminating V1 and V2 leaves Gay-Lussac's Law.

[h=2]IDEAL GAS LAW[/h]

The ideal gas law is used to approximate the behavior of a gas at conditions given by the pressure, temperature, and volume variables. Typically, the approximation is reasonable for situations close to STP (1 atm pressure/273.15 K), but deviates greatly at extreme pressures and temperatures.
Unlike the previously mentioned gas laws, there is no "initial" and "final" or "before/after" context to a problem that uses this law. Generally, this law is utilized for gas stoichiometry problems or situations where most conditions of a gas are known except for one. Let's look at a problem that involves the second scenario.
While there are five variables in the ideal gas law, one of them is the gas law constant R. It is not going to be a variable you will ever solve for, but it is a constant whose value - 0.0821 L × atm × mol-1 × K-1 - will always be plugged in for R. The unit for R looks intimidating but it is only that long because it incorporates the units of the other four variables in the problem: pressure, volume, moles, and temperature. Be aware that the unit for R dictates that any temperature substituted for T must be kelvin and any pressure substituted for P must be in atm. Consider the following problem:
Determine the volume of 45.9 g of neon gas at 78.2°C and 184 kPa pressure.

First, we know the ideal gas law must be used here because a pressure, temperature, and mass (which can be easily converted to moles) are given. Additionally, the problem asks for volume to be determined. Since P, V, n, and T are all present, the ideal gas law is the only option. Next, make sure that any values given are in the necessary units. As will often be the case, conversions will be necessary:


Now that all the conversions have been made, substitutions into the equation can be made:


[h=2]DALTON'S LAW OF PARTIAL PRESSURES[/h]

Like the ideal gas law, Dalton's law makes some key assumptions. Namely, the gases must be unreactive and follow ideal gas behavior. The law says that the total pressure of a gas mixture is equal to the sum of the pressures of each individual gas.

[h=2]DENSITY OF GASES[/h]In the derivation below, M represents the molar mass for the particular gas and m represents the mass of the gas sample. Note that unlike Boyle's, Charles's, or Gay-Lussac's Law, the identity of the gas makes a difference when determining density, but ultimately the mass of the sample does not. The initial substitution of n (moles) for m/M reflects how the number of moles of a substance is calculated - from dividing mass by molar mass.


Densities of liquids and solids are typicalls expressed in g/mL or g/cm3. However, since a mL of gas would contain a very small amount of mass, the denisty of a gas is generally expressed in g/L.

[h=2]VAN DER WAALS CONSTANTS[/h]The van der Waals constants for a gas are used when the ideal gas law is not going to give a good approximation. This happens the further one deviates from STP, and with an increased presence of intermolecular forces between particles of a gas. The van der Waals equation is given below:


This can be seen as a modification of the ideal gas law, notably with the addition of variables "a" and "b." These variables are unique for each gas and provide a calculation that is more representative (but still not always correct) of a gas at the conditions plugged in for pressure, volume, or temperature.
 

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Am I the only one lost at this point? :D

*these darn rocket scientists...
 

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Yes — but that equation implies the gas “moles” are constant — whereas in the case of altitude, the Oxygen molecules are thinner, and that in itself, would require a greater air-flow to re-produce the same result — e.g. the amount of Oxygen (not just air) passing through the relevant aperture producing the same “bang”.

From the (basic) article The Effect of Altitude on a Car or Truck Engine | Speedy Daddy

“DOES TURBOCHARGING HELP?
Yes, forced induction (turbocharging and supercharging) does help. In each of these applications the air is forced into the engine at higher pressures, effectively negating some of the loss from altitude. However, turbocharged (and supercharged) engines still lose about 1.5 percent of their power for every 1,000 feet of altitude gained. Private pilots who read the site will correctly state that turbochargers are very effective at reducing power loss at higher altitudes, but aircraft engines are specifically designed to work best at cruising levels – not sea level, as are passenger vehicles.”

This may be more relevant to fuel injection models — yet the principal is the same. Removing any restriction to air intake will help at altitude.

Now — all of this is my “gut feeling” on the subject — and Fred — please don’t get upset about something I may well have got wrong here — I’m just flying on instinct — nothing more ………
 
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