allen Posted October 10, 2002 Share Posted October 10, 2002 Mole, yes - we poor buggers work shifts and the bridge operates virtually 24/7. ☹️ It is not a public bridge, so not listed, but it is calibrated to the same standards by SCC weights & measures dept. Bridge is run by Wansbrough Paper Mill. regards allen Link to comment Share on other sites More sharing options...
JP Posted October 10, 2002 Share Posted October 10, 2002 PC, Cheers, slightly reassuring that I haven't forgotten everything. It's obviously an approximation. For the purposes of obtaining a ball park figure for power at the wheels is it reasonable to assume constant acceleration between 0 and say 60? Obviously to assume constant acc between say, 80 and 150 is plainly wrong, but how significant are rolling resistance, drag etc up to 60mph? Also how much do they vary over that range? Link to comment Share on other sites More sharing options...
JAG Posted October 11, 2002 Share Posted October 11, 2002 Ah but acceleration will not be constant! If plotted against time and produced by a single gear (no changing up) from a standing start the acceleration profile will closely match the profile of the engine power curve, but offset - the offset is losses such as rolling resistance and aerodynamic drag. this device measures acceleration and the period for which that acceleration acts. It then performs a simple calculation for Kinetic Energy; Kinetic Energy = 1/2 . M . v^2 With v = velocity (not speed) which is derived from the acceleration/time. The device knows the period during which the engine generated this energy therefore by combining the two it can calculate the power output of the engine. It knows the peak acceleration figure and will provide the peak power reading. If it had the necessary internal sophistication it could plot a graph of power against time. Although obviously not power output relative to engine rev's. All the derivation in the world isn't going to make this sound clever, I could say 'integral' this and 'differentiate' that but the end result is the same. Rate of change of Kinetic Energy is the simplest, most accurate and reliable way. With less derivation and opportunity for error. Link to comment Share on other sites More sharing options...
Peter Carmichael Posted October 11, 2002 Share Posted October 11, 2002 Am I missing something 🤔 It is power we are talking about, right 🤔 You know, the stuff that is a rate of change of energy. However we get to it, the very stuff that you emphasise as being the principle quantity measured - "the rate of change of kinetic energy" - is the answer. Anybody else coming up with the answer by any other means is still coming up with "the rate of change of kinetic energy". So did you have a point 🤔 You gave us a nice equation for what kinetic energy is, but resorted to arm-waving to explain how the machine works out the rate of change of that quantity. You sort of said it measures rate of change of energy, but then you told us it did so by measuring acceleration, working out speed (sorry, velocity, which IIRC is just speed with a sign in front of it and unless I am mistaken, the sign is thrown away in the calculation. Shall we make it easier and just use speed then 🤔). Seems you were bitching at me for using the word "integral" and then you vaguely pointed at the same answer. In case anybody didn't spot it, JAG is right that it is simple. Power is by definition a rate of change of energy and the only energy being considered is kinetic energy in a single axis. A computer that samples acceleration many times a second and determines the speed change that takes place in each sample period can keep a running total of the speed. In case of confusion, that could be termed *doing an integral*. By my reckoning this means the computer has recorded an acceleration value in each sample period and has kept a running total of the speed. In each sample period, it has the speed and acceleration values readily to hand and is able to perform calculations with them. By my reckoning the power (or, rate of change of kinetic energy) is then given by: mass x acceleration x speed. I think I wrote that earlier. Next. Link to comment Share on other sites More sharing options...
scooby dooby doo Posted October 13, 2002 Share Posted October 13, 2002 Peter, I know I'm over simplifying but I was working on linear drag with velocity for lamina flow (ie up to no more than walking pace, if that i guess) then squared above that. Is there a regime change that takes us to a cubed regime 🤔 How / why 🤔. Crudely we get turbulent at a reynolds number of 2000 ish... Simple experiments I've done with small (few cm sized) things demonstrated linear then squared. However, I didn't go above a few tens of mph. The small object means I'd only be at a pretty small reynolds number anyway. I can't remember what the kinematic viscosity of air is to work out the values... HOOPY Today I shall be mostly wanting to go for a blat R706KGU Link to comment Share on other sites More sharing options...
Micky Posted October 13, 2002 Share Posted October 13, 2002 PC wrote: "> Aerodynamic drag *power* is mostly related to the cube of road speed, not squared,<" "> Drag depends on the density of the air, the square of the velocity, the air's viscosity and compressibility, the size and shape of the body, and the body's inclination to the flow. In general, the dependence on body shape, inclination, air viscosity, and compressibility is very complex ......<" according to NASA. Acceleration can be constant (unlikely with a car), rate of change of accel is termed "jerk". Link to comment Share on other sites More sharing options...
scooby dooby doo Posted October 13, 2002 Share Posted October 13, 2002 The quote from NASA is about a particular regime. Aerodynamics is horrendously complex and drag certainly isn't with velocity squared at all times. I've seen a curve of drag against Reynolds number (Re = density x speed x 'characteristic size over viscosity and so rakes into account many of the parameters in a dimensionless fashion. it is therefore easily scalable) and it has many strange twists and turns. I'm pretty sure this curve cannot be derived theoretically and to calculate it by CFD or similar, whilst possible, is not easy... HOOPY Membership Number 4136 R706KGU Link to comment Share on other sites More sharing options...
Peter Carmichael Posted October 14, 2002 Share Posted October 14, 2002 Drag *force* is largely dependent on speed squared Drag *power* is largely dependent on speed cubed Indeed the aerodynamic drag power equation for car speeds is usually wrapped up as Cd x A x v^3 Link to comment Share on other sites More sharing options...
scooby dooby doo Posted October 14, 2002 Share Posted October 14, 2002 ah ❗ I was working in terms of force and then converting to power. I guess when i said "drag" goes with velocity squared I was meaning drag *force* and assuming everybody else would. I'm happy now I remember the first year degree question on the power from a windmill - that gave vel cubed as its *power*. Slight aside - when will manufacturers stop quoting Cd instead of Cd x A. Comments like the Merc S-Class is more aerodynamic than a VW Polo are really a bit silly but happen all the time as the Cd for each will be 0.3 and 0.33 or something. But Cd x A will be (very roughly) double for Merc.... HOOPY Membership Number 4136 R706KGU Link to comment Share on other sites More sharing options...
JAG Posted October 14, 2002 Share Posted October 14, 2002 What a load of pseudo intellectual rubbish! COMPLEXITY/COMPLEXITY and more COMPLEXITY. It's the refuge of the 'no-nothing' who's out to impress the less well educated bystander -IMHO. Link to comment Share on other sites More sharing options...
scooby dooby doo Posted October 14, 2002 Share Posted October 14, 2002 I (and I think Peter as well) like playing with maths/physics/engineering. Sometimes the results are potentially useful to other sevens, somethimes not... Or are you referring to Reynolds number and other dimensionless numbers being rubbish 🤔 Its worth noting that i did nominially the same course (24 lectures or so) at university as given by the chemical engineering dept and then by the physics dept - the overlap was less than 10%... On cam7 we were discussing particles physics a week or so back because some of us were interested (and several university degrees gave a fair bit of actual scientific fact (as near as particle physics gets to have actual facts) ). No use at all to seveners in that, but enough of us enjoyed it.... HOOPY Membership Number 4136 R706KGU Link to comment Share on other sites More sharing options...
JAG Posted October 14, 2002 Share Posted October 14, 2002 Hoopy, no problem with any of the content. It's all good stuff but has it's time and place like all other information The original posting asked for the potential influence of any error in his mass value - no body has answered him directly! Everyone (me too I guess) seems to be too busy displaying their intelect to help the poor bloke out. I have a mechanical engineering degree and work in the automotive profession doing vehicle performance predictions. I prefer to simplify things unless someone asks a direct question. Some folk seem over keen to complicate things - I presume that it inflates their ego or provides some other form of peer approval that they need! JAG Link to comment Share on other sites More sharing options...
Peter Carmichael Posted October 14, 2002 Share Posted October 14, 2002 For a look at the output of the AP-22 accelerometer clickhere You need to read the text to see that the AP-22 takes input of aerodynamic values for the losses and that in this example they had them wrong. The graph should show the same engine power in third and fourth gears if the aerodynamics were included correctly. Link to comment Share on other sites More sharing options...
Mole Posted October 14, 2002 Author Share Posted October 14, 2002 I've been following this thread with great interest and hoping that at some point someone would say something that would enable me to work out the answer to the original question - i.e. how sensitive is the power equation to inaccuracy in the mass - i.e. if I get the mass wrong by (say) 10kg - is this going to make any significant difference to the bhp figure calculated by the unit? JAG - thanks for pointing this out - even I had lost the plot a bit SV 52 CAT - the Mole is flying Link to comment Share on other sites More sharing options...
JAG Posted October 14, 2002 Share Posted October 14, 2002 Mole, I guess the answer is that it will have an effect. The magnitude of that effect is linear. If the mass/weight is 10% out then so will the answer be. The weight is used to calculate the energy output, if the weight is too high the unit will calculate a higher energy output than actual. The weight is not used for any further maths and will not have any further direct influence. JAG Link to comment Share on other sites More sharing options...
scooby dooby doo Posted October 14, 2002 Share Posted October 14, 2002 ah ❗ sorry, I see what you mean JAG. When the answer was given it was buried in a load of other stuff and not specifically pointed out. Point taken HOOPY Membership Number 4136 R706KGU Link to comment Share on other sites More sharing options...
Mole Posted October 14, 2002 Author Share Posted October 14, 2002 Thanks JAG. On this basis, if my estimate of weight is out by +/- 10kg this equates to +/- 1.4% approx (assuming all up weight of 700kg). 1.4% error seems OK given all the other factors affecting the result. Given that the bhp figure given by the unit is only accurate to about +/- 2.5%, an additional 1.4% is tolerable. Thanks all - please feel free to keep chatting among yourselves 😬 SV 52 CAT - the Mole is flying Link to comment Share on other sites More sharing options...
Micky Posted October 14, 2002 Share Posted October 14, 2002 So is it squared or is it cubed? I turned to NASA for more info and the following formula was produced as exhibit A: drag = {coefficient * density*velocity^2*reference area}/2 but no mention about power or force, so that probably explains why NASA have yet to reach Mars with a manned mission, their drag calcs are wrong. The Apollo programme (assuming it actually reached the moon) was run on the right lines, minimal electronics but maximum power........ Link to comment Share on other sites More sharing options...
Peter Carmichael Posted October 14, 2002 Share Posted October 14, 2002 The NASA equation is for drag force. When designing cars you don't need to worry too much about air density. Link to comment Share on other sites More sharing options...
Micky Posted October 14, 2002 Share Posted October 14, 2002 Thanks PC, I'll make sure that NASA are aware of that and, hopefully, they will be more specific about the drag calcs in future Micky Link to comment Share on other sites More sharing options...
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