XML Brake TEMP?

[WarpD 824]

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.

[Guest jahman]

>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.

[Engjell Berisha 1]

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!

[Guest jahman]

>please do us all the favor and give me something better than>Ed put up or give some prove to what you speak.>See: http://en.wikipedia.org/wiki/Specific_heatCheers,jahman.

[martinboehme 496]

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

[mgh 89]

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.

[mgh 89]

"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.

[martinboehme 496]

>"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

[mgh 89]

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

[WarpD 824]

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.

[mgh 89]

You lose your bet. When a body slides over another the motion is resisted by friction and the force acting parallel to the surfaces in contact, which will cause sliding is known as the coefficient of friction. If R is the mutual normal force between the surfaces and and F is the force of limiting friction then F/R = nu and nu is called the coefficient of friction. There is static friction when F is the friction force just before the suraces start to move and dynamic friction when F is the friction force when the surfaces are moving.Incidentally, it's nothing like the coefficient of lift.But I realise that you won't see this because you're staying away from forums that contain behaviour such as I'm seeing in this thread - by which I preime you mean fourms that contain posts that are clearly wrong and show a lack of understanding of basic physics.By the way, anyone considering using your formulas should think again. According to them the temperature depends directly on aircraft speed. At 250kt they show that the temperature would be 108 degC. The temperature would fall as the aircraft slowed and at 0kt, when the aircraft has come to rest, the temperature would be 0 degC ie freezing. Somehow I don't think that is correct.

[n4gix 4,949]

>But I realise that you won't see this because you're staying>away from forums that contain behaviour such as I'm seeing>in this thread - by which I preime you mean fourms that>contain posts that are clearly wrong and show a lack of>understanding of basic physics.No, he means posts like your's that merely serve to denigrate other's posts, without offering one thing concrete towards helping the original poster's question.In short, you're simply being contentious and argumentative and are not advancing any solution of your own. That's simply rude and boorish.Note carefully that I'm not contesting the validity of the points you've raised, only the method of delivery as well as the complete lack of anything truly helpful in the way of providing a solution to the question!

[mgh 89]

My original post in this thread was factual and impersonal. It pointed out that the WarpD's equations were simply wrong on a number of counts - as you appear to accept. I suggest that it was helpful to point this out, if only to avoid others being misled and going down the wrong route. Should it have been left uncorrected?WarpD's response to my original post was:"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."Is that not contentious, and argumentative, simply rude and boorish?

[n4gix 4,949]

Interesting that I find the same formula Ed cited in this discussion in a Physics forum:http://www.physicsforums.com/archive/index.php/t-50668.htmlA more robust explication on the subject may be found here:A brake disc's temperature estimation modulehttp://www.scs.org/getDoc.cfm?id=2350I would posit that a simpler guestimation would be to use the formula for KE (Kinetic Energy) to calculate the instantaneous heat rise in joules, then convert that result to a Celsius temperature/number of braking wheels.KE = (

[Guest jahman]

Bill,>to calculate the instantaneous heat riseNote that heat generation is not instantaneous, rather heat is generated as the speed decays. Heat generation is actually proportional to the square of the speed (so at half your touch-down speed you will be generating one quarter the heat per second that you generated at TD).To get temperature you have to "integrate" (sum) heat generated over time, much like to get altitude you integrate vertical speed over time.Additionally, as your brake assembly heats-up you could have the brake coefficient decrease according to a pre-determined curve, including all the way down to zero for simulated brake failure due to temperature.>Converting joules to celsius, we get 4777