Recent content by tduell

Good. It has been the 'bible' for many involved in racing for some time. It should answer most of your questions.

Have a look at "Race Car Vehicle Dynamics" by Milliken and Milliken, if you can. It gives an excellent treatment of the your problem area.

I think you are mistaken. I reckon the "pull" due to gravity is the M*g*sin(theta) (theta is angle to horizontal) and M=the total mass of the car. I reckon it doesn't make any difference where your centre of mass is located. That's how I see it, and hopefully others will correct me if I'm wrong...

Can you please explain the physics involved in your proposal? Cheers, Terry

Lateral acceleration is V^2/R. Now you have to decide on an appropriate value for lateral acceleration, and unless you make quite a few simplifying assumptions it can all become exceedingly complex. Have a look at "Race car vehicle dynamics" by Milliken and Milliken to get a better understanding...

Well, I think the torque required depends on the angular acceleration you want for the gun. Torque = I * alpha, where I = mass moment of inertia and alpha = angular acceleration (consistent units) You should be able to find formulas for mass moments for common shapes and axis of rotation by...

OK. Could you please provide the "why is it so"? Cheers, Terry

I think that even in the non rigid container case the pressure will always settle at the vapour pressure for the temperature. Cheers, Terry

I think the vapour pressure depends only on the temperature, so my guess is that the fluid level won't have any effect. Cheers, Terry

angular deflection = (M*pi*D*N)/(E*I) where M = applied moment, or torque D = mean coil diameter N = number of active coils E = young's modulua I = second moment of area of wire Use consistent units. Cheers, Terry

I'm not familiar with doing all the symbolic maths in a post, so I'll have a go with ancient methods. If you know the ships rotational inertia (I) about the axis if rotation, then you can use T=I*alpha, where T=applied torque (the thruster or whatever), and alpha is the angular acceleration. Use...

It is expanded in the high pressure turbine to an intermediate pressure, then passed back to the boiler where it is reheated at constant pressure, then expanded in the low pressure turbine. I would guess the 'procedure' to only have partial expansion is really related to the design of the...

I think the most common arrangement would be a small centrifugal pump driven by an electric motor. If it isn't working, there are a number of possibilities... (1) the water in or out is blocked (2) the electric motor has failed It is really a matter of checking the simplest/easy things...

You are quite right, well spotted as they say. That had slipped by me. It is a real good reason to try to mount the struts close to vertical. If one was sufficiently motivated, one could write an analysis package in Octave, or whatever, with a little database of strut specs, and run a trial...

OK, looking at this arrangement, when closed, the angle between the strut axis and the vertical is 81 deg, and the vertical component of the strut force is 0.15 (cos 81), hence total strut force required is 60/0.15=400, or 200 lbf per strut. When open the angle between strut axis and vertical...