Recent content by Gordon Arnaut

Thanks, Tim. I'm glad I asked. I actually have a solution to the problem and that is to use curves with second degree polynomials. I can just cut the airfoil into smaller pieces and use curves defined by three points instead of four. The result is a function that looks like this: y...

What about expressing the curve in terms of parameters? Then we can use the formula for parametric equations: L = ∫ba √(dx/dt)^2+(dy/dt)^2 dt

Thanks again Tim. I put in those missing superscripts. It's looking better now. I hate to give up so quickly. What about integrating by parts? That way you could have a quadratic function in each part?

Thanks Tiny-Tim. I went back and put in those characters. Why do you say there is not an exact way of integrating this expression? Can you explain? Regards, Gordon.

Homework Statement I'm trying to compute the circumference of a wing section. I have broken up the airfoil circumference into arc pieces and used cubic splines to come up with an equation for each piece. For example, the arc nearest the leading edge of the wing is the function: y =...

No, this is not an easy problem. The tire friction was not given, but this is probably something that could be looked up. I'm not sure it would come into play that much anyway. Let's take a similar problem, where you have a fast-moving vehicle that hits a boulder and sends it flying...

I wonder if someone might suggest an approach to solve this problem? A vehicle weighing 5000 lbs is moving at 10 mph and is struck from behind by a vehicle weighing 5000 lbs and moving at a high rate of speed. The slow-moving vehicle is pushed forward a distance of 150 feet. How fast was...

Okay thanks, Berkeman. Actually there are more data points, so it becomes more like a curve than two straight lines. I'm going to try function transformations to see if I can come up with a formula that comes close to matching the curve. Regards, Gordon.

Berkeman, The relation describes a curve, not a line. See the attached graph. Regards, Gordon

How can I derive the equation from a Graph? The x coordinates are: 28,30,32,33,34,35 The Y coordinates are: 7,8,9,10,11,12 We could express this as a relation, r: r{(7,28), (8,30), (9,32), (10,33), (11,34) (12,35)} Regards, Gordon,

Thanks, Andrew. You're right. The answer I got is actually correct. What threw me was that in Otto cycle mathematical models, the temperature of combustion mentioned is always a lot lower. However, if you take the high temperature number I calculated above and multiuply it by the...

Andrew, it's just simple artihmetic using hte values I provided. We don't know the cv post-combustion because we don't know the temperature. Regards, Gordon.

I'm trying to estimate the temperature of combustion in an Otto Cycle engine, using a formula based on the First Law: T4 = T3 + fQ /cv Where T4 is the combustion temperature, T3 is the air temperature after compression, f is the fuel/air ratio, Q is the heat energy of the fuel, and cv is...

Thanks, Q. What is the underlying equation? Regards, Gordon.

Sorry, forgot to mention the gas is air. Yeah, I can get Cv once I know Cp, by subtracting R: Cv = Cp - R. I was just wondering if there was an equation because I want to use it in a spreadsheet. Regards, Gordon.