Recent content by 5.0stang

Hi guys/gals, I had a couple random questions on leverage and the human body. I am an active individual that enjoys lifting weights and working out. At the gym today, I got to thinking about the "mechanical advantages" that certain body types have doing particular lifts. First question...

Okay I got another one that I need help with. Maybe someone can help explain it to me. :) Formula on paper looks like: (Pi*d^2/4) * 1-((cos(Open Angle + Closed Angle)) / (cos(Closed Angle))) I am using this for a throttle body throat area formula. D = Minimum Diameter and I am using...

Okay, guys, resurrecting an old thread. How can one find the area of a D shaped port? Multiply the long side of the D, and the width, and then multiply by .75?

I asked this question back in January, but I do not mind that it was brought back-up. I just purchased Gordon Blair's book (D&S of a 4-stroke engine), and it goes into a lot of detail, that is hard for me to pick-up. You guys put it in easier terms for me to understand. David - Yes, the...

I have read that hot gases flow better than cold gases. Why is this so? Less dense? Any good readings on it. I am thinking of incoming intake charge (cool) in an engine, and the outgoing (hot) gases. Thanks.

Would not the air behind the ball (stagnated) be of a higher pressure. Thinking Bernoulli's principle here...

Sorry for the late response. I figured out the problem. The professor made a typo in our question, and she corrected it. I solved it very quickly after that. Thanks anyways!

Nevermind.

It sure does, thanks!

Okay, the formula that I am working from is: (RL+S) – sqrt [ RL^2 – (S*sin CA)^2 ] – S*cos CA Rod Length = RL = 5.090 Half Stroke = S = 1.5 Crank Angle = CA = 107.5 With numbers plugged in (my version?): (5.090+1.5) - sqrt [5.090^2 - (1.5*sin (107.5))^2] - 1.5*cos (107.5) The...

With components, why is it best to knife-edge the backside? For example: Knife-edge a crankshaft Knife-edge the back of a throttle body blade What principle is this following?

I am trying to solve for volume. I have 4m ^2 given in the problem. Okay, the formula to work from is: 1. Fb = (Density in kg/m^3)(Volume)(Acc. Due to Gravity) 2. 519.4 = (1000 kg/m^3)(V)(9.8 m/s^2) 3. 519.4 = 9800V 4. Volume = 18.87 (units?) Is this correct so far? How do I...

Okay, just looking over my notes one more time. My teacher said to use the weight of the girl (53kg) to equal Fb. How do I convert 53kg to Newtons? I can use online calculators, but that is not helping me learn. It looks like it is 9.8 x 53 = 519.4 Newtons So I should be getting Fb =...

[SOLVED] Basic Buoyant Force HW Question First time doing this, so bare with me:) Homework Statement A 53kg girl dives off a raft 4m square floating in a freshwater lake. By how much does the raft rise? Density of water = 1000kg/m^3 Use the equation for buoyant force. Then solve...

I notice consistently in the aftermarket exhaust world that in the majority of cases, when a longtube is placed on an engine the low to mid-range power picks up. This is also with the same size diameter primaries but different lengths. So I am comparing: Long tube: 30" primary tubes...