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I have a floor pump intended for mainly rather low pressure tires.

It lets out quite a bit of air from the hose when disconnected.

Can I calculate (approximately) how much air pressure is lost by measuring the hose length and diameter?

I need to get it up to 100psi, so should I pump it to 110psi, 120psi, 130psi?

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    Unless you're pumping up tires for a race car, you don't need to be exact (tire pressures aren't an exact science) - just play around with the level you pump it to until you get good results (which is supposed to be how you set your tire pressures anyway). The amount of air let out depends on how quickly you remove the chuck, which varies decently from time to time.
    – Batman
    Commented Apr 19, 2014 at 15:21
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    Uh, i was under the impression that the most of the air left out comes from the hose not the tube itself. Roadbikes have rather stringent pressure requirements in order to prevent flats. I can't leave the flat-prevention aspect of correct pressure to feeling alone.
    – JBeurer
    Commented Apr 19, 2014 at 15:35
  • You should start by calibrating your pump. You can't improve the precision, but you can improve the accuracy. Commented Apr 19, 2014 at 15:58
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    The air in the hose is irrelevant. It's the amount of air that sneaks past the tire valve before it snaps shut that is the issue. And this is in large part a matter of technique. Commented Apr 19, 2014 at 17:32
  • I use a Presta to Schrader adapter for my presta tubes. This allows the pressure alone to open the valve and no air leaks at all when removing the pump head.
    – Chris
    Commented Jun 8, 2017 at 17:09

2 Answers 2

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Can I calculate (approximately) how much air pressure is lost by measuring the hose length and diameter?

No, you cannot tell how much pressure is lost based on the size of the hose. This is not because there is not enough information to tell but because the hose is irrelevant.

You seem to be under the impression that there is some total amount of pressure in the hose and in the inner tube, and when you disconnect the pump and release the pressure in the hose, some of it is lost.

This is an incorrect understanding of what pressure means. Pressure is an intensive property — it is something that is the same for of all the air no matter how you divide it up within the connected system. The pressure in the hose equals the pressure in the inner tube, which equals the pressure in the whole pump-and-hose-and-inner-tube system.

In the tradition of physics education, let's consider the ideal case with a separately controllable valve on the tube. This is what would happen:

  1. With the pump attached to the inner tube, you open the valve. This causes the pressure in the hose and tube to equalize: this is when the gauge on your pump jumps upward. At this point, the pressure in the tube has reduced, because some of the mass of air in it moved into the hose to pressurize it.

  2. You use the pump to increase the pressure as desired, according to the gauge.

  3. You close the valve on the tube. The pressures on each side of the valve remain equal to each other and to that read by the pump's gauge.

  4. You disconnect the hose. The extra air in the hose escapes to the atmosphere.

As soon as you closed the valve, the inner tube became a closed system: the pressure in it is no longer affected by anything about the pump. It will stay where the gauge said it was while you were pumping.

Now, in a real valve and pump chuck, there isn't a separate control to close the valve (pumps may have moving center pins to push the valve open as part of the “lock” but they aren't always exactly the right length — but I've never used a Presta valve so I may not have this right for those), and it is instead closed as you pull the chuck off the valve. This may release some air from the tube, due to the valve not being fully closed before the connection starts being open to the atmosphere. However, the amount of air released from the tube depends on the valve and how fast you let it close. It does not depend on the volume of air in the hose (except insofar as the pressure in the hose opposes the pressure of the tube even though both are also flowing out to the atmosphere, which is probably a minor effect).

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    Okay, I'm missing something here and it's at step 3. How would I close a Presta valve before disconnecting the hose? Commented Apr 20, 2014 at 4:31
  • @CareyGregory - The valve shuts automatically when disconnecting the hose. The nut that you can tighten is just a "safety device" that ensure the valve stays shut... or am I misunderstanding your question?
    – mac
    Commented Apr 20, 2014 at 13:24
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    I think point 3 is misleading: the valve is not manually shut, it shut by itself when the external pressure drops below the internal one (i.e.: when the hose is removed).
    – mac
    Commented Apr 20, 2014 at 13:26
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    @CareyGregory Yes, real inner tube valves aren't separate. I cover that in the last paragraph. I've edited to emphasize that the first description is idealized.
    – Kevin Reid
    Commented Apr 20, 2014 at 13:38
  • @mac and Kevin - Yeah, I understand all that but when you disconnect a pump from a valve it's not an instantaneous thing. Some air is lost from the tube during the disconnect procedure. The amount lost depends on how long long the tube's valve was left partially open while the pump was being disconnected. I can, for example, leave a tire almost flat by pumping it to full pressure and then removing the hose very slowly. I take the OP's question to be how should he measure that amount? Commented Apr 21, 2014 at 4:16
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If you're that concerned with lost air and exact pressures, get Schrader valves on your bike and buy a shock pump like the Scott Shock Pump Pro 600, which lets you close the valve before disconnecting the hose.

Other than that - if you're not far heavier than 100kg, it doesn't really matter whether you have six or eight or twelve bar in the tubes when it comes to puncture safety. In those pressure ranges it's just a matter of rolling resistance and comfort. The former optimizes somewhere around 8bar, the latter decreases steadily with increasing pressure.

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