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mgh

XML Brake TEMP?

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Does it even exist I've been looking for the code for ever and I still got nothing. I'm just trying to get a string to show me brake temperature, thats all, nothing more but IDK how to do it in any other way then XML.

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I don't think there's a variable in FS that provides brake temp.

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Anyway I would go about this? Is there a wheel temp variable or something related to the brakes that would give me a temp close to the brake temp.

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Perhaps you could measure the time brake is applied and wheel rotation ,and link that to an animated needle or a dialface , putting a fair bit of lag on the rotational animation , if you were to link this node to another that read the OAT it would baseline there.basicly you fake it , mind you i am thinking modelled gauges in the vcand you are talking string values for a gauge

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No there is no wheel temp var in FSX, a reasonable thing because it's not something that could be simulated in a generic form.Does your aircraft measure direct temp or instead an index like 0 to 5, as some airliners use (ie 757/767) ?Tom

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heat=fpV, where f is friction coef, p is pressure, and v is velocity.

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Does that mean that I can get it to give me a brake temp reading? And if so how I didn't quite understand what you said mate.

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That's the formula... it's basic physics. It's not part of FS, as I said FS doesn't supply brake temp.Pick a friction coefficient of a known brake pad in the real world (just find one on the internet)... determine braking pressure based on max hydraulic pressure versus amount of brake applied (just a value to use)... get velocity from ground speed. That should give you something to work with.

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Assuming the following:f = friction coefficient of the brakes = 0.50p = pressure of the brakes = 3200 psiV = velocity = 250kts250kts = 128.6 meters per second0.50 * 3200 * 128.6 = 205760 joules205760 joules = 108 degrees C = 227 degrees FThat make it clearer?

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Awesome mate!!!Now can I put that into code so that flight simulator could do the formula and give me a figure on the brake temp? And if I did the Friction Coefficient of the brakes and the Pressure would be constants right? The only thing that I would put as a variable for FS to understand would be the V and it would plug that in with my GS?

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The formulas are wrong.0.50 * 3200 * 128.6 = 205760 joulesGiven that the friction coefficient (0.5) is dimensionless the LHS of your equation is:pounds/square inch * Metres/secThat is an incompatible mixture of Imperial and SI units and doesn't give Joules which are Newton*Metres and represent work/energy/quantity of heat. Even converting pound/sqare inch to SI units (Newtons/sq metre) the LHS becomesNewtons/sq Metre) * (Metres/sec) = Newtons/(Metre*Sec)which is still not Joules.205760 joules = 108 degrees C = 227 degrees FTemperature is different unit to a Joule and there is is no universal relationship between them.

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>The formulas are wrong.>>0.50 * 3200 * 128.6 = 205760 joules>>Given that the friction coefficient (0.5) is dimensionless the>LHS of your equation is:>>pounds/square inch * Metres/sec>>That is an incompatible mixture of Imperial and SI units and>doesn't give Joules which are Newton*Metres and represent>work/energy/quantity of heat. Even converting pound/sqare inch>to SI units (Newtons/sq metre) the LHS becomes>>Newtons/sq Metre) * (Metres/sec) = Newtons/(Metre*Sec)>>which is still not Joules.>>205760 joules = 108 degrees C = 227 degrees F>>Temperature is different unit to a Joule and there is is no>universal relationship between them. To be blunt, you've just jumped into something as if you're the world's foremost expert on the subject. Well, you're not... and you're wrong. End discussion. Seriously.

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> End discussion. Seriously.Erm... why? He has a number of good points.First of all, your equation is for power, not heat (which is power times time). That's why the units don't work out.Second, the "pressure" in your equation is kind of misleading, because "pressure" in physics is usually force per unit area; what you mean is the force applied to the brake pads. That, multiplied by the coefficient of friction, gives us the retarding force. That in turn, multiplied by the velocity (of the brake disc going past the brake pad) gives us "power" (i.e. the amount of kinetic energy we're dissipating per second), as per your equation. We would have to multiply this with the length of time that the brakes were applied to get the amount of energy dissipated, or heat.Third problem is, this amount of energy still doesn't tell us how hot the brakes are going to get. First of all, this would depend on the specific heat capacity and the mass of the components that we're heating; additionally, we're losing a lot of heat to the environment, and this effect depends on the speed of the air going past the brakes. I would say that it's very hard to model / calculate all of this accurately from first principles... I would guess that addons that simulate brake temperatures do this using data that was measured on the real aircraft.Cheers,Martin

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Might I suggest that too much effort has been invested in the straining of gnats?Don't overwork the physics; look at the desired output and work with the cause and effect relationship. When the cold and dark aircraft is sitting on the ramp, the brake temperature is the ambient temperature as shown on the OAT gauge. Let the code change the temp variable by a small percentage each time it refreshes. Adjust the variable pos/neg value and the gauge refresh rate so the temp increases slowly whenever brakes are applied and decreases slowly (but never below ambient) when brakes are off. If you want to get fancy, introduce an aircraft speed variable to the process. Test-taxi the aircraft (including aborted takeoffs) and adjust values as necessary.In the end, you'll have a gauge or gauges that follow the dynamics of the brake heating/cooling cycle. It won't be perfectly accurate, but sometimes we have to accept that FS9 cannot model reality to the nth degree.Cheers,Glenn

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Ok... here's what I did:I went to a friend who teaches physics at a local college. I discussed the question with said professor. Said professor stated that there are lots of variables in calculating the temperature... however... said professor gave me the formula I provided and stated it would emulate rather accurately the heat produced from the friction.To prove the concept, consider that the friction coefficient drops with heat build up (brake fade)... as the coefficient drops, the resulting value rises. It works very much like the real brakes do with regards to temperature. It also is supposed to represent an instantaneous moment... it's up to the user to develop it further, i.e. calculate heat retention vs dispersion. Unless I'm expected to write all the code as well?FYI: Joules is a measure of energy... it is not limited to a specific type of energy. Heat is a form of energy. There is indeed a correlation. In fact, there are formula to convert joules to celcius (temperature). Imagine that! ;)Some of you jump into discussions with merely a desire to rip a post to shreds. You really don't stop to think: "Did the answer provide what the poster requested?"Now, if you want to argue it further... go to a college and argue it. I at least went to a source of knowledge.

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>In fact, there are formula to convert>joules to celcius (temperature). Imagine that! ;)There is *one* formula, and it requires you to know the specific heat of the element that will register the temperature:Change in Temp = Change in Heat * Specific Heat * MassTo complicate matters, as temperature increases, cooling increases due to:1. Convection (cooling proportional to temperature difference), and2. Radiation (cooling proportional to absolute temperature).So actually brake temp is a fairly complex variable to model.>Some of you jump into discussions with merely a desire to rip>a post to shreds. You really don't stop to think: "Did the>answer provide what the poster requested?"Maybe some jump-in because they know physics? ;-)Cheers,jahman.

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As far as that goes logically you'd think that joules being a form of energy itself could be translated into heat also a form of energy.Any way what I ask for was brake temp help not a arguement. And if you're going to try and discredit someone like that please do us all the favor and give me something better than Ed put up or give some prove to what you speak.Ed thanks so much mate you've been a big help, now I don't really know what to do with the formula I was wondering if it is possible to get flight simulator to do the formula. Also I suspect the answer would need to be added to the ambient temperature because when I used the formula with your figures and a velocity of 12 knots I got a temp of 6C. Anyway I traced the company that makes C-17 brakes and emailed them I may call em tomorrow asking for the pressure of the brake and the friction coefficient.Anyway thanks once again Ed you've been a great help!!!2nd Lt. Engjell BerishaGood Day!

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Ed, no desire to rip your post to shreds here... I very much value and respect the contributions that you make here and the time and energy you put into doing so. I simply wanted to add something to your answer because IMO it didn't provide all of what the poster requested. I realize that my answer was probably too terse to be useful to Engjell; I'll try to correct that now.I'll clear up a small point first to avoid confusion later on:> As far as that goes logically you'd think that joules being a form of energy itself could be translated into heat also a form of energy.When you say "heat" here, do you mean something that is measured in Fahrenheit or Celsius? A potential source of confusion here is that the everyday meaning of "heat" (a temperature measured in Fahrenheit or Celsius) is slightly different from what physicists mean when they say "heat" (a certain amount of energy). In the following, I'll exclusively use "temperature" (measured in Fahrenheit or Celsius) for the former and "heat" (measured in Joules) for the latter.To illustrate the difference: Say you have a gram of water and you add a certain amount of "heat" to it, say 4.2 Joules (maybe by holding a candle underneath it for a certain amount of time). This will increase the temperature of the water by one degree Celsius, so if its temperature was 20 degrees before, it's now 21 degrees. In other words, the specific heat of water is 4.2 Joules per gram per Kelvin ("Kelvin" is just a fancy way of saying "a temperature difference of one degree Celsius").Now, say we take ten grams of water and again add 4.2 Joules of heat to it. Because we're putting the same amount of heat in but have more water to heat, the temperature won't rise as much... in fact, because we have ten times as much water, the temperature will only rise by a tenth of what it did before, i.e. by 0.1 of a degree. So if the water temperature was 20 degrees before, it will now be 20.1 degrees.So there is a way of converting heat to temperature, but to do so we need to know the mass of the object that we're heating and what it's made of (which will tell us the specific heat).OK, getting back to the subject of brakes: How do we work out how much heat is going into the brakes? Ed's formula tells us how many Joules of heat are being put into the brakes per second. (This amount of heat transfer per unit of time is called "power" and is measured in Joules per second, or Watts.) Here's the formula, I've merely changed it to use standard variable names:P=u F vwhere P is the power, u is the coefficient of friction, F is the force with which the brake pad is being pressed onto the brake disc, and v is the velocity with which the brake disc is going past the brake pad. (Note: This is usually not the same as the velocity of the aircraft... if the diameter of the brake disc is a quarter of the diameter of the wheel, then the brake disc is going past the brake pad at a quarter of the ground speed of the aircraft.)So just to make things more concrete, let's plug in some numbers here, which I'm taking out of thin air. Let's say u=0.5, F=1000 N, and v=20 m/s. Then we get:P = 0.5 * 1000 N * 20 m/s = 10000 W(Newtons times metres are Joules, and Joules per second are Watts.)Let's say this amount of braking is applied for one second -- how much heat will that put into the brakes? Well, 10000 W times one second is 10000 Joules. (We have to choose the interval of time short enough that the velocity doesn't change a lot during that time interval, so maybe a tenth of a second would be better in a real implementation.)OK, so now we know we're putting 10000 J of heat into the brakes. To simplify things, we'll say that we're heating only the brake disc, no other component of the system. Again pulling numbers out of thin air, we'll assume that the brake disc weighs one kilogram, and that's it's made of steel, which has a specific heat of about 0.5 J / g / K. So we need 0.5 Joules to heat one gram of steel by one degree Celsius, or 500 Joules to heat the whole brake disc (which weighs 1000 grams) by one degree Celsius. Since we've put in 10000 Joules, we've heated the brake disc by 20 degrees Celsius during the one second that we applied the brakes. So say we had a "brake temperature" variable, and it was set to 15 degrees before, we would increase it by 20 degrees to get 35 degrees.So much for putting heat into the brakes. Now how about the heat dissipated by the brakes due to the air flow going past them? Unfortunately, this is where it gets complicated... see here for the whole story:http://en.wikipedia.org/wiki/Convective_heat_transferMy feeling is that it would be very hard to quantify all of the properties of the system to work out accurately how much heat you're losing due to convection. Instead, I would suggest doing the following:P_convection = c1 * (T_brake - T_air) * (v_air + c2)where P_convection is the amount of heat you're losing per second, T_brake is the temperature of the brake, T_air is the temperature of the air going past the brake, v_air is the speed of the air going past the brake, and c1 and c2 are constants that you'll have to fudge until you get results that look right. (This is what I meant about it being hard to calculate these kinds of things from first principles...) Intuitively, what c1 models is the area of brake disc that is exposed to the air -- the greater it is, the more heat you'll lose. c2 models the fact that, even when the aircraft is standing still, you'll still be losing heat at a certain rate because the heated air will rise, causing an airflow past the brakes (this is what the Wikipedia article means by "natural convective heat transfer").Given all of this, say P_convection works out to 5000 W, and let's say we're again looking at a time interval of one second. Then within that one second, the brakes would lose 5000 J of heat, and we could again use the specific heat, as above, to work out how much this would reduce the temperature of the brakes.Phew.Of course, given that we have two constants whose values we have to "fudge" anyway, there's no real point in working out what the actual force applied to the brakes is, what the weight and specific heat of the brake disc and any other components of the brake are, etc., because we would only be modelling half of the system realistically. So this takes us back to what Glenn suggested: Use these, or similar, formulas to give you the right qualitative behaviour, and then tune the constants until the values you get "look right" or agree with what you see on the real aircraft.Hope this helps,Martin

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I am not wrong. You clearly don't understand the difference between tthe various physical units and you equations, to be blunt, are simply nonsense and blustering doesn't change that fact.

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"There is indeed a correlation. In fact, there are formula to convert joules to celcius (temperatureI'd be interested to see the formula - especially as temperature and heat are two totally different things physically. I "jumped in" because the your formulae, being totally wrong, don't give the OP what he wants.

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>"There is indeed a correlation. In fact, there are formula>to convert joules to celcius (temperature>>I'd be interested to see the formula - especially as>temperature and heat are two totally different things>physically. Hey, easy... no need to beat a dead horse, right? ;)Martin

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I only just got back to this thread. Anyway, I'm not sure the horse knows it's dead!

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Out of curiosity, anyone know what the purpose of the friction coefficient actually is? I'm willing to bet you don't. I'll give you a hint... it's very much like the lift coefficient for a wing in it's origin and purpose.Oh, and I am done. This will be my last post in this forum. I have a rule of staying away from forums that contain behavior such as I'm seeing in this thread.Good day.

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