tag:blogger.com,1999:blog-1564958057737541664.post8395900916473667509..comments2015-03-06T05:43:12.083-08:00Comments on On Bicycles, and.... what else is there?: Bernoulli, stagnation pressure, and barometric altimetersdjconnelhttp://www.blogger.com/profile/01484858820878605035noreply@blogger.comBlogger7125tag:blogger.com,1999:blog-1564958057737541664.post-76387673731950962702010-02-11T06:06:05.208-08:002010-02-11T06:06:05.208-08:00This comment has been removed by a blog administrator.如此的http://www.blogger.com/profile/11914259155278125948noreply@blogger.comtag:blogger.com,1999:blog-1564958057737541664.post-43617634989229523302010-02-09T07:48:54.959-08:002010-02-09T07:48:54.959-08:00Ha! See?..that's why you're obviously a m...Ha! See?..that's why you're obviously a much smarter guy than I am. I didn't even think about that rho cancellation/simplification...duh. Nice. I'm going to modify that old spreadsheet and see what alpha works best after that change.<br /><br />I think the confusion came into this from my terminology about what % of free stream velocity was "presented" to the sensor. You were looking at it as what amount was actually being stopped at the sensor, while I was referring to how much was flowing past the sensor (i.e. the % of free stream was what had NOT been stopped by the "bluff body").TnAhttp://www.blogger.com/profile/08175472546482777614noreply@blogger.comtag:blogger.com,1999:blog-1564958057737541664.post-6353547067569350392010-02-09T06:48:46.775-08:002010-02-09T06:48:46.775-08:00I think the conversion you did is a bit indirect: ...I think the conversion you did is a bit indirect: an unnecessary linearization.<br /><br />Think about it this way: the pressure at a given altitude = the weight of all air in a column above that altitude. So if I change altitude by Δz, the change in pressure is the change in weight of the column. That's the negative of the weight-density of the air multiplied by the height difference, where weight = gravity multiplied by mass, so if I know the mass density ρ (mass per unit volume = mass of a gas molecular multiplied by molecules per unit volume), I multiply by g and Δz and take the negative, and that gives me the change in pressure with height. That's how the altimeter measures height (for ρ, it uses the ideal gas law, so it needs T, also).<br /><br />So then I consider the dynamic pressure. That's just from Newton's law: pressure = kinetic energy of the scattered air per unit are = ½ ρ v².<br /><br />So if I then calculate the equivalent altitude change from this "dynamic" pressure, I the ρ cancels. I get:<br /><br />Δz = ½ v² / g<br /><br />Very simple. No rules of thumb for pressure change with altitude required.djconnelhttp://www.blogger.com/profile/01484858820878605035noreply@blogger.comtag:blogger.com,1999:blog-1564958057737541664.post-72302698633704771622010-02-09T00:09:26.970-08:002010-02-09T00:09:26.970-08:00Aaah...I see where I may have been unclear. When ...Aaah...I see where I may have been unclear. When I said that only 5% of the free stream velocity was present at the port, that means that 95% has been "converted" to higher pressure. I used an "alpha" of .95 in my spreadsheet.<br /><br />Also, I'm not sure about your equation for "delta z"...I used the equation that the stagnation pressure(in Pascals) was = 1/2*rho*(alpha*V)^2. I then converted the stagnation pressure to equivalent elevation difference, which is basically linear at low elevations (IIRC it was 1 millibar per 8 meters...but I'm having a hard time recreating the altitude conversion factor I used in the spreadsheet) Make sense?TnAhttp://www.blogger.com/profile/08175472546482777614noreply@blogger.comtag:blogger.com,1999:blog-1564958057737541664.post-8403087188290656132010-02-07T14:56:21.589-08:002010-02-07T14:56:21.589-08:00Thanks! I fixed that typo: I'd meant 15 mps, ...Thanks! I fixed <a href="http://www.biketechreview.com/forum/viewtopic.php?f=1&t=1826&sid=fb6b10a4a87eac695a677108bd4033f3#p25952" rel="nofollow">that typo</a>: I'd meant 15 mps, not 15 kph. My fingers were typing without neural engagement, as is their regular habit. So the post was correct with this fix: the numbers come out too low relative to the plot.<br /><br />Good point on the stagnation bubble! That makes sense. Although then the question arises to what a theoretical value for the velocity (or dynamic pressure) multiplier is (the pressure multiplier the square of the velocity multiplier). With your "bubble" concept, it stands to reason this number might be fairly large.djconnelhttp://www.blogger.com/profile/01484858820878605035noreply@blogger.comtag:blogger.com,1999:blog-1564958057737541664.post-13521019900856315652010-02-07T14:55:45.296-08:002010-02-07T14:55:45.296-08:00This comment has been removed by the author.djconnelhttp://www.blogger.com/profile/01484858820878605035noreply@blogger.comtag:blogger.com,1999:blog-1564958057737541664.post-17854924461178466732010-02-06T10:00:23.374-08:002010-02-06T10:00:23.374-08:00Good stuff again, Dan. But, you have to remember ...Good stuff again, Dan. But, you have to remember that if the sensor is mounted anywhere near the front of the rider/bike system, you'll see an elevated pressure due to the "stagnation bubble" caused by the rider. So, even if the sensor port itself is oriented perpendicular to the general flow, you're still going to see an elevated pressure.<br /><br />Oh...and I saw your post over on BTR. I think the "mistake" is your assumption of the speed when the max delta elevation was seen...it was more like 15 m/s, or ~54 km/hr at those points.TnAhttp://www.blogger.com/profile/08175472546482777614noreply@blogger.com