# Testing the parasitic bike pump

Inspired by the parasitic bike pump we posted about last week, Simon Jansen decided to try it out for himself. Using his own car tire to avoid the moral issue of borrowing air from an unsuspecting motorist, and a sweet looking motorbike that he is building as the test subject, he found an interesting result– the pressure on his bike tire always ended up lower than that of the car:

One thing that annoyed me was people seem to think that a car tyre inflated to 30PSI would be able to then in turn inflate the bike tyre to 30PSI. That’s obviously wrong as the air has to come from one tyre (at 30PSI) to the other flat tyre (at 0PSI) so you’ll end up with them both at some pressure in between. I assumed they’d both end up at the same pressure.

In the end I decided the easiest way to see if this works was to try it!

So, here is what I did. I inflated the car tyre to 30PSI. Seems like as good a pressure as any and representative of what you’d get on a normal car. I made sure the minibike tyre was fully deflated – no pressure on the gauge. I hooked one end of the hose onto the mini bike then I hooked the other end onto the car tyre.

What happened next is the air could be heard hissing down the hose and the bike tyre would inflate. I would wait a while till the hissing stopped (5-10 seconds) then quickly remove both valves at the same time.

He repeated this experiment many times, using various auto tire pressures, and measured the resulting pressure at the end of each experiment. The curious part is that the minibike tire always ended up at a lower pressure than the automobile tire.

My guess is that both tires lose a fixed volume of gas when you remove the valves, and that this affects the pressure of the smaller bike tire more than the car, however we aren’t sure. Anyone know what’s happening?

### 30 Responses to Testing the parasitic bike pump

1. Clau Alex on said:

it’s because of the lower inner surface area of the bike tire, combined with the fact that the bike tire is a narrow cylinder with less of a load in pounds per sqare inch from the low weight of the bike that would result in geometrical deformation. therefore the cars’s bigger tire, and it’s square-ish transversal section, along with the weight of the car, of whom about a quarter is distributed on a surface a few inch square, leads to inner deformation of the chamber and thus to reduction of available volume relative to the volume of the gas, in this case, air. and you get higher pressure in the car (tyre)

• The weight/load of the car on the tire would not have that effect otherwise tire pressures would drop when you jack a car up, which is not the case.

2. https://me.yahoo.com/a/fozTYVUxkdjfAvdrKbA.7wUXNKlWlBzF#7493e on said:

Apologies if I am missing something, but Boyle’s law states that for a fixed temperature, the Pressure * Volume of gas is constant.

We are increasing the volume of the gas here, by having the same gas fill both tires. Therefore the pressure drops in both. This would certainly be the case if the donor tire were not mounted on a car.

There may be a second order effect of changing the size of the car tire due to the weight of the car pressing down on the tire, slightly changing it’s volume, but to maintain the pressure at exactly the same level, the product of the 2 volumes must remain constant.

3. whitcwa3579545 on said:

Two possible reasons:
1. He didn’t wait long enough for the pressure to equalize. Just because he didn’t hear air flowing doesn’t mean it wasn’t.
2. Disconnecting the hose from the bike caused enough air to escape to affect the reading.

• Using a ‘T,’ connect a pressure gauge to the tube between the two tires. When the reading stops changing, remove one end of the tube, check reading again. Replace end, check reading, remove other end, check reading, replace end, check reading final time. Report back!

4. try leaving it connected overnight and see if that gives the pressure time to equalize. Initially, when there is a large difference in pressures the air will flow quickly, but as the difference in pressure decreases, so will the rate of flow.

5. Matt,
Now I want to know about the Cafe Racer next to the MG?

6. Alan Blue on said:

One of the areas of contention in the last thread was about how much pressure is going down in the car’s tires. It would be nice if the actual numbers were posted here instead of just “The bike’s tire’s pressure was always lower.”

So, if the car’s tire was at 30psi and the bike’s tire is flat, how low precisely did the car’s tire pressure go?

7. The pressure difference is coming from the actual hose itself. He said he tried to remove the hoses at the same time. However statistically looking at his data, doing this he isn’t in control. This means that the way he did this contributed to the error that he saw. But the actual pressure drop is because when disconnecting the hose, one side had to provide the volume of air in the hose; which escaped. So once again looking at his data it shows that he initially was removing the bike side first, then the car side. And then midway through his test he started taking it off the car first.

So this means that in the beginning the car tire was providing the volume to the hose, and later the bicycle was providing the volume to the hose.

So your intuition is right on. The car tire, possessing a higher volume, will be less affected by the small drop in volume from the hose, while the bicycle tire will be affected much more.

So what should have been done was only remove one side of the hose. Not to mention make it a ONE WAY valve on the bicycle end. this would mean that when filling from a car tire, it is ensured that the car provides the volume of air in the hose that is lost, not the bike.

8. keithwins on said:

Schrader tire valve hoses/connectors have a spring-forced closure mechanism built in, and the tire pressure has to overcome that spring pressure for air to pass. So the net pressure in the 2nd tire will be the original pressure minus that spring pressure (accounting for the area on which the spring is acting, and the change in net volume from one tire to two).

A simple test to see if this is correct: have two similar tires (easier but not necessary). Do the parasitic transfer. Then, change the order of filled/empty tires. I think the pressure differential will come out the same, proving that it’s the transfer system (the hoses & connectors) that are creating the pressure differential, not anything having to do with the tires (though the actual volumes of the two tires do pertain to the final average pressure, of course — which is why similar tires will be easier to get good readings from than if you try to fill a car tire with a bike tire).

Of course, the 2nd tire doesn’t need to be empty, just lower enough in pressure (greater than the hose/connector delta-t) that air moves in the direction you want.

• The springs/valves in the Schrader valve won’t affect it. The fittings on air pumps/bike pumps have a little nub that presses the valve down so that air may be pressed in. So the air pressure won’t for all intensive purposes interact with the spring.

9. bored_engineer on said:

The flow rate depends on the pressure difference in the two vessels. I don’t know how to write math in these comments, but looking at the equation in front of me, the only variable in the flow rate that we can’t neglect is the pressure differential. (This is an equation for compressible flow through a pipe. If you examine Bernoulli’s principle, you can see this too.)

This can be tested if you have an air compressor:
1. Attach a hose and the sprayer nozzle that came with your compressor.
2. With the pressure regulator turned so that the exiting pressure is as high as it will go (probably 120 psi), observe the effects. The air will exit with high velocity, making lots of sound and disturbing the surroundings.
3. Now turn the pressure regulator down as low as it will go, so that it reads about 5 psi. You’ll notice that the velocity of the exiting air will go down considerably, disturbing the surroundings much less, and with much less noise.

10. crrieger on said:

All these responses and no one has bothered to mention Robert Boyle and his law – P1 x V1 = P2 x V2. Assume P1 is the initial pressure of just the car tire and V1 is the initial volume of the just the car tire. P2 would be the final pressure of the system (car tire and bike tire) and V2 would be the final volume of the system. Flow rate between the tires doesn’t matter much here since we would let the system go to equilibrium.

What this means – P2 can never be higher than the pressure of the car tire and the bike tire will still be only filled around 1/3 of the way assuming it needs at least 100 psi. Since pressure and volume are inversely related, any small change in volume will result in a much larger change in pressure proportionally for the bike tire since it has far less volume than the car tire.

11. Yes, it’s true that the rate of exchange of air will be a function of the difference of pressure between the two tires (probably a non-linear function if the flow is turbulent), but my intuition tells me that the pressure in the two tires will equalize (to within the resolution of your pressure guage) in a fairly short amount of time.

My best guess is it has to do with air loss during removal of the hoses from the valves, and also air loss during the measurement of the pressure (every pressure guage I’ve ever used always looses a bit of air when you use it).

To eliminate this discrepancy, attach an extra valve to each tire and mount a pressure guage to each new valve. This way you can always see the pressure in each tire without having to remove hoses. Make sure the two guages are calibrated to each other.

Then deflate the bike tire and attach the hose between the two original tires. You can watch the pressure rise on the bike tire and the pressure fall on the car tire. Unless my understanding of Pascal’s Law is wrong, they should eventually equalize. Then when you remove the hose between the two tires, you can observe how much the pressure in each tire decreases due to escaped air during hose removal.

12. whitcwa3579545 on said:

He removed the hose as soon as the hissing stopped. Just because the sound stopped doesn’t mean the flow had stopped.

13. keithwins on said:

NOT the Schrader valve. In many Schrader fill connectors (sorry, I don’t know what the proper name is), there is a small spring to keep the connector closed when there’s no air pressure.

The simple proof would be whether he can simply blow through his apparatus when it’s not connected to any schrader valves. I bet not. Even more realistically, connect a schrader valve to each end (sans tires) and try.

Of course, we were never given the magnitude of the discrepency, so it could also be that a certain amount of air leaks upon pressure testing, and that affects the pressure of the small tire more. This could be confirmed by testing repeatedly: the difference should grow.

14. I thought this would cause some comments! Remember, I wasn’t trying to do a rigorous, controlled scientific experiment or statistical analysis. I just made the thing as it was originally described and used it as you would use it ‘in the field’ as it were. There are lots of ways this could be improved of course. And the device itself could be improved as well (as I, and others noted, a one way valve would help a lot).

To clear up a few things I was removing both valves at the same time as quickly as I could. You have to since as soon as you remove one end air will leak out of the tube of course. There isn’t any kind of valve in the connectors I was using to close them when they aren’t connected. I don’t know if you can get connectors like that that do? With the hose disconnected you can blow right though it.

Also I would wait for the hissing to stop before moving the tubes. I was doing this in a quiet garage on a Sunday sitting right near it. My impression was that a few seconds after the you can’t hear anything anymore the pressures are not going to be changing greatly. You can definitely hear the flow of air slowing when you hook this up. I don’t believe it would take long for the pressure to reach its final point. A matter of a few seconds I would think. Would need to measure that to know for sure. A rough estimate of the time from attaching the hose to pulling it off again would be about 10 seconds. Maybe that’s not long enough?

The car tyre pressure isn’t affected much by measuring the pressure. If your pressure gauge is dropping a few PSI from the tyre every time you try to use the gauge you need a new gauge! The bike tyre will be affected more since it’s volume is less but still not enough to matter after one reading I think. I was careful to only take one reading though and wrote down that one just in case.

Also the tyre pressure gauge I use is not exactly a highly calibrated, scientific instrument so my measurements aren’t that great. I do have a better gauge but that’s being used on another project just now.

Someone can probably come up with a much better experiment to show how this all works but I think the original aim was achieved which was to show how this device, made as per the Instructable, would actually work. You can transfer air from a car tyre to a flat bike tyre. But both will end up at a lower pressure than the starting pressure. And that pressure is probably too low for a bike anyway.

Now, as to the mini-bike that I might have to explain later but basically it was a little project I started at night school years ago (about 8) back when the government here funded such things. It ran once then the clutch on the little 2 stroke I was using burned out and I never got around to finishing it! I have a few old engines about at home. I should look into finishing it off one of these days. It was really an excuse to practice pipe bending and my gas brazing skills and do some fibre-glassing. The tank and mud guards were made from that.

Simon

15. I think keithwins is right, there seems to be a one way valve that keeps the handle of the pump from coming up with the pressure in the tyre and allows you to fill the tyre. Tom

16. whitcwa3579545 on said:

There is a one way valve in the tire, but it is kept open by the stub in the clamp-on hose connection. Hand pumps also have a one way valve to prevent the tire from inflating the pump. So when I connect my hand pump to my tire I can read the tire pressure on the pump’s gauge, but no air flows until the pump pressure exceeds the tire pressure.

17. Dustbuster7000 on said:

OK, my long comment got eaten. The answer is probably flow losses in the small diameter connecting hose. Try a longer hose and see if the final pressure difference is greater between the two tyres. Friction losses will be a reasonably large component of the losses in the flow, but they’ll be easier to see if you do your experiements at higher pressure (start with the car tyre at 40psi and run 5 experiements with two or three difference hose lengths).

• Dustbuster7000 on said:

Oh, and losses through all the fittings will contribute something to the difference as well.

• keithwins on said:

Well, if there’s no spring-loaded valve in the apparatus (you say you can blow through it), I think it must be the differential effect of leakage at the time of disconnect.

Regardless of how much internal friction there is in the apparatus, the two pressures will equalize eventually: probably fairly quickly. When taking off the fittings, assuming the same volume of air leaked from each tire at the same time (that is, they both start at the same pressure and we’ll assume you remove the connectors in the same length of time), then that leaked volume of air would differentially influence the pressure in the (smaller-volume) bike tire.

Best test would be my earlier suggestion of trying the experiment with two equal-volume tires, I think.

18. This is probably due to the effect demonstrated in the 5th grade science project where two inflated balloons are connected by a simple valve. When the valve is opened, the more inflated balloon will actually suck most of the air out of the smaller one, instead of equalizing. Not sure exactly why this happens, But I remember it has something to do with the ratio of surface area to volume.

19. You guys are over complicating this way too much.

The pressure difference is due to the hose.

Take Boyles law:

P1*V1=P2*V2

P1 is the pressure of the Tire initially at 30PSI, and V1 is the volume of the system initially (just the tire). V2 is the combined volume of the system, hose, bike tire, and car tire.

If you do some rough calculations you can see that the volume of the car tire is roughly 10:1 ratio to the bike tire, and that the hose is 1:10 ratio of the bike tire.

So knowing that lets do some calculations.

Pressure of the system after connection:

P1*V1=P2*V2
(30PSI * 100in^3 )/111 in^3 ~= 27.0

This shows that after we connect our system it will be at roughly 27PSI for all components.

So now we need to see what happens to the pressure if the bike tire or car tire is supplying the volume to the hose that is lost.

First the car tire:
V1 is 100, and V2 is the hose and car tire volume 101.

P1*V1=P2*V2
(27PSI * 100in^3 )/101 in^3 ~= 26.75

This is a delta of ~0.25 PSI which his gage probably wouldn’t be able to resolve.

Next the bike tire:
V1 is 10, and V2 is the hose and bike tire volume 11.

P1*V1=P2*V2
(27PSI * 10in^3 )/11 in^3 ~= 24.6

This is a delta of ~2.5 PSI which is almost identical to his average delta.

So just by doing these simple calculations we can rule out that the small “spurt” of air released when measuring is negligible as well as any flow restrictions. And we can then assume that the bulk of the volume loss is due to the hose.

20. Dustbuster7000 on said:

Capn,
You calculation are right but your assumptions are wrong. First, you’ve already accounted for all of the volume in the new system (car tyre + hose + bike tyre) in the first formula, which gives you 27psi. More importantly, you assume that all of the air lost from the system when the hose is removed will come from one tyre. There no reason to suppose that is the case.

If we assume a frictionless system, everything in the system (car tyre, hose, bike tyre) will be at a uniform 27 psi before the hose is removed. If no air leaves the system when the hose is removed (and Simon said he removed the hose from both tyre at the same time, so that’s a reasonable bet), then both tyres will be at the same pressure when their respective valves close; the pressure they were before the hose was removed. The hose vents its pressure to atmosphere. Which should result in both tyres being the same pressure. But they’re not.

keithwins idea is interesting, since both valves are the same type and both tyres are at the same pressure, each might leak the same additional (small) volume of air as the hose is removed. And that volume would be a higher proportion of the bike tyre volume, resulting in a lower final pressure for the bike tyre. I just don’t know if that loss would be significant enough to cause the difference Simon observed.

I hope Simon is interested in doing more experiements to resolve this, but regardless, its a fun discussion

• No my assumptions are right. The system, as in the bike tire, hose, and car tire are at a uniform 27 PSI when it reached equilibrium.

In the “real world” there is no such thing as “simultaneous”. When he “simultaneously” removed the hose the bike tire, or the car tire has an extra volume attached to it. And perhaps in rare cases the bike tire and car tire contributed the same amount of volume to the hose. This is where we get the variation from.

Your assumptions are wrong however. The bike tire, nor the car tire can “close” its valve during fill because they are schrader valves. The valve is fixed open by the connection on the hose. This inevitably links the hose to one of the tires as an extra volume. Now if he had a valve as close as possible to the connection and closed those off after the tires reached equilibrium, you’d have an argument. However that is not the case.

• Dustbuster7000 on said:

I agree that simultaneous is probably too strong, but close enough for our purposes. We already know that this isn’t a precision scientific experiment, Simon says as much.

The valves *will* close when the hose is removed. That is the point when air can leave the system. And the air in the hose is not an extra volume, you’ve already accounted for it in your calculations (100+10+1).

If we say Simon removes the car tyre slightly ahead of the bike tyre, *then* the system is two volumes (bike tyre and hose), each at 27 psi, connected at the bike tyre valve. Then a fraction of a second later you have two separate volumes each at 27 psi. Then the hose vents its air to atmosphere and the bike tyre valve closes and it reamins at 27 psi. Now if air moves from the bike tyre to the hose (or to the atmosphere) between those two moment, before the valve can fully close, then we have what keithwins was describing, a leak of more air during the hose removal.

21. As one gas decompresses and the gas in the smaller tube compress you loose thermal energy.

The pressure drop is due to this difference. The car tire is loosing pressure and temperature at the same time. The bike tire is gaining pressure and increasing in temperature at an higher rate.

Less the stretch and expansion in the transfer medium and the pressure gauge. (remember measuring something changes what you measure)

If you waited long enough for the pressure and temperature gradients to balance then you would see the equalization you are all looking for.

-Cecil

22. Billy Boberson the Third on said:

Look up head loss. It’s mostly applied to liquid flows, but the concept is the same for gaseous flows as well