For Questions That Don't Merit Their Own Thread

[munlander1]

8 hours ago, Wjolcz said:

Why do big galaxies appear globular? Is it because they have their disc hidden and closer to the core?

An example galaxy? Are you talking about elliptical galaxies?

 

 

Are spacecrafts "bigger" in space because you aren't bound to the floor like you are on earth?

[Exploro]

I'm trying to construct a model to estimate the initial length of an elastic band to be used for a engineering design group project vehicle.

I've found a way that gives me the change in length and thought of using a re-written form of Hooke's Law would help find Li.

Hooke Law is expressed as F = kx, where F is equal to the spring constant of the material times it's displacement. Hooke's law is related to Young's Modulus were F = E*Area/Initial Length of Material. So I re-wrote Hooke's Law substituting the aforementioned  expression thinking I could eliminate force as a variable. But I think I made a mistake. For According to Khan Academy k = E*A/Li. So I end up in a situation where Li = Li. So I've clear in error.

Anyway my questions are: Am I going in the right direction investigating either Hooke's Law or Young's Modulus to find initial length? If so, how can I do so when for the moment the force required to distort the band is not known?

[Wjolcz]

11 hours ago, munlander1 said:

An example galaxy? Are you talking about elliptical galaxies?

IC 1101

[Racescort666]

11 hours ago, Exploro said:

I'm trying to construct a model to estimate the initial length of an elastic band to be used for a engineering design group project vehicle.

I've found a way that gives me the change in length and thought of using a re-written form of Hooke's Law would help find Li.

Hooke Law is expressed as F = kx, where F is equal to the spring constant of the material times it's displacement. Hooke's law is related to Young's Modulus were F = E*Area/Initial Length of Material. So I re-wrote Hooke's Law substituting the aforementioned  expression thinking I could eliminate force as a variable. But I think I made a mistake. For According to Khan Academy k = E*A/Li. So I end up in a situation where Li = Li. So I've clear in error.

Anyway my questions are: Am I going in the right direction investigating either Hooke's Law or Young's Modulus to find initial length? If so, how can I do so when for the moment the force required to distort the band is not known?

Assuming your knowns are F, x, E, A and you're trying to find L_i, L_i = (E*A*x)/F

[munlander1]

7 hours ago, Wjolcz said:

IC 1101

I'm pretty sure that multiple collisions with other galaxies will slowly mold the combination into a more elliptical structure.   

[RealKerbal3x]

The questions/answers on this thread tells me that this is definitely the most intelligent gaming forum on the internet

Anyway, my question: why exactly did the Falcon Heavy core crash? I’ve heard that it ran out of fuel and that the engines didn’t restart. What actually happened?

[DerekL1963]

31 minutes ago, RealKerbal3x said:

Anyway, my question: why exactly did the Falcon Heavy core crash? I’ve heard that it ran out of fuel and that the engines didn’t restart. What actually happened?

Nobody outside of SpaceX knows exactly what happened, and inside SpaceX they're under NDAs.   The announced cause is that they ran out of igniter fluid.  I suspect further discussion belongs in the SpaceX fanboy thread.

[Racescort666]

@RealKerbal3x to build on what @DerekL1963 has said, they use TEA-TEB injected into the engine which automatically combusts with oxygen (even liquid oxygen) to start the engine. You can even see when they inject it during engine starts because it’s burns green. They also used it to start the engines in the SR-71 and A-12. Think of it like starting fluid for your lawnmower/snowblower/whateverelse is hard to start. (Now that I think of it, it’s probably closer to ether start on a diesel but nobody deals with that much anymore so let’s not go there.)

During the landing, the center cluster engine did fire (as can be seen in the video) but one or both of the side engines were not lit. If the announcement is taken at face value, one or both of these engines was either completely out or didn’t have enough starting compound to ignite. Speculation on which case or other causes we will leave to the fan forum.

[RealKerbal3x]

15 hours ago, DerekL1963 said:

Nobody outside of SpaceX knows exactly what happened, and inside SpaceX they're under NDAs.   The announced cause is that they ran out of igniter fluid.  I suspect further discussion belongs in the SpaceX fanboy thread.

 

7 hours ago, Racescort666 said:

@RealKerbal3x to build on what @DerekL1963 has said, they use TEA-TEB injected into the engine which automatically combusts with oxygen (even liquid oxygen) to start the engine. You can even see when they inject it during engine starts because it’s burns green. They also used it to start the engines in the SR-71 and A-12. Think of it like starting fluid for your lawnmower/snowblower/whateverelse is hard to start. (Now that I think of it, it’s probably closer to ether start on a diesel but nobody deals with that much anymore so let’s not go there.)

During the landing, the center cluster engine did fire (as can be seen in the video) but one or both of the side engines were not lit. If the announcement is taken at face value, one or both of these engines was either completely out or didn’t have enough starting compound to ignite. Speculation on which case or other causes we will leave to the fan forum.

OK, interesting. Just wondered.

[The Flying Kerbal]

Does anyone know if the Soviets give their N1/L3 Lunar Landing project a name such as "Apollo"?

Edited April 10, 2018 by The Flying Kerbal

[TheDestroyer111]

 

On 4/10/2018 at 11:47 PM, The Flying Kerbal said:

Does anyone know if the Soviets give their N1/L3 Lunar Landing project a name such as "Apollo"?

Sorry, there is no single project name other than some weird abbrevations/codes or general descriptions like "Soviet moonshot". Soviets/Russians, unlike USA, don't give official nicknames to unfinished projects and even if some military/space technology becomes operational, they often keep the older, less creative name or abbrevation.

That's how we ended up with a large number of Kosmos xyz spacecraft (or almost-spacecraft).

 

My question: Why do most boats and ships (except submarines) have a totally blunt stern? Isn't the ideal hydrodynamic shape a teardrop, ie more rounded front and finer back, just like submarines?

Edited April 15, 2018 by TheDestroyer111

[p1t1o]

50 minutes ago, TheDestroyer111 said:

My question: Why do most boats and ships (except submarines) have a totally blunt stern? Isn't the ideal hydrodynamic shape a teardrop, ie more rounded front and finer back, just like submarines?

If it were moving through air, or water, it would have a rounded stern. But a boat moves along the interface between air and water, so the optimum shape is subject to more complex factors. You will usually find though, that the part of the stern which is under the water does conform more to the familiar teardrop geometry.

Sometimes extreme hydrodynamic efficiency may also be sacrifices for things like ease of steering or better positioning of propulsion gear.

Submarines need to be quiet (inefficient shapes are noisy), and so maneuverability takes a hit for better streamlining.

 

 

[TheDestroyer111]

Then, those long and sleek competitive rowing boats have been officially tagged as REBELS and a reward shall be set up for capturing one alive and bringing it to the police station of Earthrace or average sail racing dinghies!

20 hours ago, p1t1o said:

Sometimes extreme hydrodynamic efficiency may also be sacrifices for things like ease of steering or better positioning of propulsion gear.

ikr thats prob the reason 99% time, but it's always that stupid little 1% that leads to eureka

 

Is there any known watercraft guru on this forum?

 

//EDIT:

14 hours ago, DerekL1963 said:

The body of rotation form wasn't introduced for quieting.  It was introduced for better underwater performance and to allow a single screw.  The latter was particularly important for nuclear boats as the amount of power you can put into the screw without cavitation is more-or-less proportional to the area of the blades.  Twin screw boats were limited in the size of the screw, and that sharply limited performance.

If it's proportional, then two half-size screws are equivalent to one full-size )))))

Edited April 14, 2018 by TheDestroyer111

[DerekL1963]

6 hours ago, p1t1o said:

Submarines need to be quiet (inefficient shapes are noisy), and so maneuverability takes a hit for better streamlining.

The body of rotation form wasn't introduced for quieting.  It was introduced for better underwater performance and to allow a single screw.  The latter was particularly important for nuclear boats as the amount of power you can put into the screw without cavitation is more-or-less proportional to the area of the blades.  Twin screw boats were limited in the size of the screw, and that sharply limited performance.

In fact, the body of rotation hull introduced a new form of noise - blade rate, which arose from the interaction between the wake of the sail and the rotating propeller.

[DerekL1963]

20 hours ago, TheDestroyer111 said:

If it's proportional, then two half-size screws are equivalent to one full-size ))))

The area of a circle is 2(pi)r².

Area of a circle of size 1 = 1.  Area of a circle of size 2 = 12.

So, no.  Two half size screws are not equivalent to a single full size screw - because of that exponent.

[TheDestroyer111]

14 minutes ago, DerekL1963 said:

So, no.  Two half size screws are not equivalent to a single full size screw - because of that exponent.

[insert politician saying "wrong" meme here]

In my definition, half-size screws have half area.

 

ofc I totally agree that 1 screw is harder to cavitate due to blade tip speed 

 

E: Thing is, cavitation is not so much dependent on screw area as on tip speed, that has a tendency to create low pressure spots...

This just got me a crazy idea... what if you fitted your sub with a huge screw to maximise area relative to tip speed... like wider than your boat 

Edited April 14, 2018 by TheDestroyer111

[DerekL1963]

4 minutes ago, TheDestroyer111 said:

In my definition, half-size screws have half area.

Nobody cares what your definition is.  The math shows it to be wrong.

[peadar1987]

On 13/04/2018 at 5:38 PM, TheDestroyer111 said:

My question: Why do most boats and ships (except submarines) have a totally blunt stern? Isn't the ideal hydrodynamic shape a teardrop, ie more rounded front and finer back, just like submarines?

Below the waterline, the hull does taper to a more teardrop shape. The square stern will potentially be to maximise the efficiency of the packing of the containers. A teardrop shape would potentially be more hydrodynamically efficient, but the elongated shape with parallel sections amidships gives more deck area.

For high performance watercraft, there is the potential for planing, where hydrodynamic lift raises the boat out of the water and significantly reduces drag.

Sailing dinghies have a few different design considerations. In general the waves are larger in relation to the boat, so they have a sharper, wave-piercing bow. The bow section also tends to be narrower and deeper than the stern to minimise the drag in displacement mode. The aft section tends to be broader and flatter to generate more lift and get the boat planing sooner. How you trim the boat will depend on whether you have enough power to get her planing or not.

[kerbiloid]

Probably a sharp tail would weight more than save with better aerodynamics, and make additional problems (so, efforts→time→cost) due to uneven rows of containers..

Spoiler

Reminds of
AlphaBall

 

Edited April 15, 2018 by kerbiloid

[KG3]

On ‎4‎/‎13‎/‎2018 at 12:38 PM, TheDestroyer111 said:

My question: Why do most boats and ships (except submarines) have a totally blunt stern? Isn't the ideal hydrodynamic shape a teardrop, ie more rounded front and finer back, just like submarines?

Large container ships are built with a relatively flat bottom to reduce it's draft (or depth below the waterline).  It's a tradeoff.  The more hydrodynamic the shape the deeper the draft is going to be.  Ports need to dredge channels to accommodate these bigger ships and this just gets expensive.     

[KG3]

Is there a difference between "altitude" and "distance"?  For instance Wikipedia says that:

"The Moon's average orbital distance at the present time is 384,402 km"

but Wikipedia also says that:

"A geostationary orbit can be achieved only at an altitude very close to 35,786 km"

Is there some point where altitude changes to distance, or does one refer to a natural object, or maybe one refers to an object that was manufactured on and left the surface of Earth?

[YNM]

3 minutes ago, KG3 said:

Is there a difference between "altitude" and "distance"

Vector ?

A strictly radially-away vector would count under "altitude", a distance is any vector.

The Moon is only roughly 384402 km away from me. At least it deviates by 12742 km throughout a day.

[monophonic]

1 hour ago, KG3 said:

Is there a difference between "altitude" and "distance"?

In this context of orbits around a central body the difference is where the zero point is. Altitude is calculated from the surface of the central body while distance from the center of mass. So a satellite in a geostationary orbit is at altitude of 35,786km above the Earth and distance of 42,164 km from Earth, both at the same time. The difference is the Earth's equatorial radius, 6,378km.

All numbers copied from the wikipedia article.

https://en.m.wikipedia.org/wiki/Geostationary_orbit

[sevenperforce]

22 minutes ago, monophonic said:

2 hours ago, KG3 said:

Is there a difference between "altitude" and "distance"?  For instance Wikipedia says that:

"The Moon's average orbital distance at the present time is 384,402 km"

but Wikipedia also says that:

"A geostationary orbit can be achieved only at an altitude very close to 35,786 km"

Is there some point where altitude changes to distance, or does one refer to a natural object, or maybe one refers to an object that was manufactured on and left the surface of Earth?

In this context of orbits around a central body the difference is where the zero point is. Altitude is calculated from the surface of the central body while distance from the center of mass. So a satellite in a geostationary orbit is at altitude of 35,786km above the Earth and distance of 42,164 km from Earth, both at the same time. The difference is the Earth's equatorial radius, 6,378km.

All numbers copied from the wikipedia article.

https://en.m.wikipedia.org/wiki/Geostationary_orbit

The reason for this distinction is that when you're discussing objects close to the Earth, it makes a good deal of intuitive sense to talk about their altitude, just as you would with, say, a plane. A small single-engine prop plane like a Cessna 172 flies at up to 4-5 km. A jet airliner flies at 10-11 km. Weather balloons ascend to 30-50 km, and the edge of space is recognized at 100 km. From there, it's natural to say stuff like "the Apollo missions had a parking orbit at 185 km" and "the ISS orbits 400 km above the Earth."

However, as you move outward, it gets tricky. The purpose of most MEO and HEO orbits has a lot to do with orbital periods, so you need the semi-major axis, which depends on the distance from the center of the planet rather than the distance from the surface. Thus, for GEO satellites and stuff like the moon, you'd want to report "distance from Earth", so GEO is 42,165 km distance.

Yet that gets even more complicated when we are talking about transfer orbits. Suppose you are going from a 200 km parking orbit out to GEO. You'll need to enter a Hohman transfer orbit with a perigee of 200 km altitude and an apogee of 42,165 km distance. That's confusing. So you end up reserving two different values for GEO; 35,786 km altitude and 42,164 km distance. A transfer orbit is thus described in terms of altitude (200km x 35,786km) even though you'd find the semimajor axis (and thus the period) of that ellipse by taking the actual distance of 6,578 km for the perigee with the actual distance of 42,164 km for the apogee and dividing the sum by two.

[YNM]

There is one more difference : Distance is usually centre-to-centre length, altitude is closest distance from surfaces.

So GTO is nominally 42157 km away (distance) from Earth. The Moon sits an an altitude of 378031 km.

 

It only becomes really blurry when the thing you're talking above have distances of hundreths of AU.

Edited April 21, 2018 by YNM