Community > Forum > Technology & Science > Orbital Mechanics

Orbital Mechanics

Posted by: campbelp2002 - Thu Dec 23, 2004 7:49 pm
Post new topic Reply to topic
 [ 160 posts ] 
Go to page Previous  1, 2, 3, 4, 5, 6, 7, 8 ... 11  Next
Orbital Mechanics 

Could an object spiral into the Sun?
Of course! 64%  64%  [ 16 ]
No way! 28%  28%  [ 7 ]
I used to think so, but now I don't. 0%  0%  [ 0 ]
I didn't think so before, but now I do. 8%  8%  [ 2 ]
Total votes : 25

Orbital Mechanics 
Author Message
Moon Mission Member
Moon Mission Member
User avatar
Joined: Tue Oct 05, 2004 5:38 pm
Posts: 1361
Location: Austin, Texas
Post    Posted on: Thu Jan 13, 2005 3:29 pm
Ekkehard Augustin wrote:
as far as I remeber we didn't agree that the tangential component is along the ellipse... )

and
Ekkehard Augustin wrote:
What's along the ellipse is the vector that is resulting if the vector that is tangent to the ellipse and the vector that is pointing towards the sun are calculated to each other.

Well you have contradicted yourself in this one post. How can a vector that is tangent to the ellipse not be the tangential velocity? (EDIT) by definition the tangential velocity is the one tangent to the ellipse.


Back to top
Profile WWW
Moderator
Moderator
avatar
Joined: Thu Jun 03, 2004 11:23 am
Posts: 3745
Location: Hamburg, Germany
Post    Posted on: Thu Jan 13, 2005 3:47 pm
1. The movement along the tangent is going to another direction than the movement along the ellipse ==> to take the velocity along the tangent as velocity along the ellipse could be prove to be an error.

2. The ellipse is caused by the gravitational acceleration tpwards the sun and this accelerations is causing velocity along the direction towards the sun - but Earth isn't going towards the sun.

3. If the tangential vector under 1. and the vector towards the sun under 2. are calculated by vector calculations the correct direction is got - the velocity towards that direction proves to be different to the velocity along the tangent: It's greater.

The calculations can be done by drawing the vectors but they can be done too by trigonometrics and a third alternative are calculations within a coordinate system. I am using the second alternative.



Dipl.-Volkswirt (bdvb) Augustin (Political Economist)


Back to top
Profile
Moon Mission Member
Moon Mission Member
User avatar
Joined: Tue Oct 05, 2004 5:38 pm
Posts: 1361
Location: Austin, Texas
Post    Posted on: Thu Jan 13, 2005 3:50 pm
(EDITED)
I do not understand your reply.

Does the new magenta vector in the diagram represent your idea of the "tangential" velocity?

If not, is there a vector in the diagram that DOES represent your idea of the "tangential" velocity?

If not, please describe how I should draw a “tangential” vector and I will change the magenta vector to show that.
Image


Back to top
Profile WWW
Moon Mission Member
Moon Mission Member
User avatar
Joined: Tue Oct 05, 2004 5:38 pm
Posts: 1361
Location: Austin, Texas
Post    Posted on: Thu Jan 13, 2005 4:34 pm
OK, I have been studying your reply and have some comments.
Ekkehard Augustin wrote:
to take the velocity along the tangent as velocity along the ellipse could be prove to be an error.

NO! If you don’t take the instantaneous tangential velocity as the true velocity along the ellipse you are in error.

Ekkehard Augustin wrote:
The ellipse is caused by the gravitational acceleration tpwards the sun and this accelerations is causing velocity along the direction towards the sun

NO! You have it backwards. The Sun causes the acceleration. Then the acceleration causes the ellipse.
(EDIT) OK, I changed my mind. You ARE saying the sun causes the acceleration and then the acceleration causes the ellipse. Right?

Ekkehard Augustin wrote:
If the tangential vector under 1. and the vector towards the sun under 2. are calculated by vector calculations the correct direction is got - the velocity towards that direction proves to be different to the velocity along the tangent: It's greater.

I still don’t understand this.
(EDIT) I may understand now, but it makes no sense. I have not drawn a green acceleration vector at the Sun for position 1. I can draw one if you like, and another one at position 3. These different green vectors would be the ones to use for calculating the change in vectors 1 and 3, not the one at 2.

Ekkehard Augustin wrote:
The calculations can be done by drawing the vectors but they can be done too by trigonometrics

These two things are exactly the same. You are using trigonometry to calculate the vectors.

We need to agree on the vectors before any communication between us has any meaning. All physics starts with drawing the vectors. After all vectors are drawn to represent real physical quantities then, and only then, can calculations start. Please refer to my previous post and answer the questions so we can agree on the vectors.


Back to top
Profile WWW
Moon Mission Member
Moon Mission Member
avatar
Joined: Tue Feb 10, 2004 2:56 am
Posts: 1104
Location: Georgia Tech, Atlanta, GA
Post    Posted on: Fri Jan 14, 2005 1:13 pm
I missed the first part of this conversation, and regrettably don't have the time to read it all, so campbelp2002: are you in the engineering/orbital dynamics field?

campbelp2002 wrote:
If not, please describe how I should draw a “tangential” vector and I will change the magenta vector to show that.
Image


Uh..... Last time I checked (that would be about three weeks ago), tangential vectors are specifically tangent to the ellipse. The sun's gravity will have the effects of changing the magnitude of both the tangential and radial components of the acceleration (and therefore velocity and motion), but it cannot change the direction of the tangential vector. Tangential and radial vectors exist in fixed directions, as do i, j, and k. Seems to me, you're talking about the actual acceleration vector (which is composed of both the tangential and radial components). That *can* change direction, but the individual components cannot.[/i]

_________________
American Institute of Aeronautics and Astronautics
Daniel Guggenheim School of Aerospace Engineering

In Memoriam...
Apollo I - Soyuz I - Soyuz XI - STS-51L - STS-107


Back to top
Profile
Moon Mission Member
Moon Mission Member
User avatar
Joined: Tue Oct 05, 2004 5:38 pm
Posts: 1361
Location: Austin, Texas
Post    Posted on: Fri Jan 14, 2005 1:38 pm
spacecowboy wrote:
I missed the first part of this conversation, and regrettably don't have the time to read it all, so campbelp2002: are you in the engineering/orbital dynamics field?

No. I was going to be an astronomer and got bachelor degrees in both physics and astronomy, but never completed astronomy graduate school. Now I am a UNIX system administrator. I developed my orbit simulation for a moon lander game. Here is my new Java based simulator:
http://home.austin.rr.com/campbelp/orbit.html
The start of a work in progress.

spacecowboy wrote:
Uh..... Last time I checked (that would be about three weeks ago), tangential vectors are specifically tangent to the ellipse.

YES! But Ekkehard has specifically denied that black vectors 1,2 and 3 are tangential in that diagram. I am trying to find out what he DOES consider tangential.

I am talking about the actual instantaneous velocity being modified by the actual instantaneous acceleration. The Black and red vectors in this diagram. And my simulation uses the blue and green components for calculation. The code is shown in an early post in the disposal of radioactive waste topic. (EDIT) See the top of page 3 of that topic. Not code, but an explanation. (EDIT AGAIN) Found the code near the bottom of page 4.
Image


Back to top
Profile WWW
Moderator
Moderator
avatar
Joined: Thu Jun 03, 2004 11:23 am
Posts: 3745
Location: Hamburg, Germany
Post    Posted on: Sat Jan 15, 2005 5:15 pm
Hello, Peter,

it is very difficult to answer to your questions to me because your post with the questions is a mixture of right and wrong vectors but you tend to change into better understanding of my vectors.

The tangential vector is a vector touching the orbit at only one point - this vector is representing velocity towards a direction touching the orbit but less than orbital velocity.

The vertical vector is causing accelleration which partly changes the direction the object is going to from the tangential vector into the direction along the orbit and partly increases the velocity of the object to the exact orbital velocity - this is the reason the resulting vector under 3. is increased compared to the tangential vector.



Dipl.-Volkswirt (bdvb) Augustin (Political Economist)


Back to top
Profile
Space Station Commander
Space Station Commander
User avatar
Joined: Mon Oct 06, 2003 9:22 pm
Posts: 844
Location: New York, NY
Post    Posted on: Sat Jan 15, 2005 9:59 pm
i just read the last page of this, and it makes no sense at all. why don't you guys just use symbols to define your vectors and stop trying to figure out what terminology the other guy is using.

ps, from what i did understand, it seems that peter is right as far as terminology goes.

_________________
Cornell 2010- Applied and Engineering Physics

Software Developer

Also, check out my fractals


Back to top
Profile
Moon Mission Member
Moon Mission Member
User avatar
Joined: Tue Oct 05, 2004 5:38 pm
Posts: 1361
Location: Austin, Texas
Post    Posted on: Sun Jan 16, 2005 2:19 am
Ekkehard Augustin wrote:
The tangential vector is a vector touching the orbit at only one point - this vector is representing velocity towards a direction touching the orbit

So you agree that black vectors 1,2 and 3 are the tangential vectors?
Image
Ekkehard Augustin wrote:
but less than orbital velocity.

But you don't think these tangential vectors represent the tangential velocity?

Do I understand you corectly?


Back to top
Profile WWW
Moderator
Moderator
avatar
Joined: Thu Jun 03, 2004 11:23 am
Posts: 3745
Location: Hamburg, Germany
Post    Posted on: Sun Jan 16, 2005 4:25 pm
Peter,

yes, your black vectors in the last posted diagram seem to be tangential vectors. Please extend them backwards to make clear that they are embedded into a real tangent.

This diagram of yours is an example and so I don't know what km/s they are representing - this makes it a little bit to answer your last question. I can only answer like this: tangential vectors always do represent tangential velocities. But tangential velocities don't be orbital velocities because a tangent is a straight line whereas an orbit is a curved line - and such acurved line in reality can be caused only by at least one additional velocity towards a different direction than the direction of the tangent.

Hello, TerraMrs,

I know mathematical symbols but don't be used to make use of them - so to use them might cause additional misunderstandings and I might use them wrong by random often. This would be complicating all which I want to avoid. My terminology is a humble translation from German - in German I would call the "tangential vector" "tangentialer Vektor". I don't know another name for it right now. I could call it in German "Tangente" but this term wouldn't include a restricted lenght representing velocity.

Peter's diagrams are helpful but obviously don't be representing Earth's orbit but an ellipse in general whereas I use a circle as an approximation of Earth's orbit - what I never would do in the case of Pluto for example.



Dipl.-Volkswirt (bdvb) Augustin (Political Economist)


Back to top
Profile
Moon Mission Member
Moon Mission Member
User avatar
Joined: Tue Oct 05, 2004 5:38 pm
Posts: 1361
Location: Austin, Texas
Post    Posted on: Sun Jan 16, 2005 7:18 pm
OK! We agree the 3 black vectors are tangential velocities. Of course there are an infinite number of different tangential vectors. I have only shown these three for illustration.

I disagree that the tangential velocity is different from orbital velocity. However, for now I will not argue that. Instead, I want to better understand how you think of orbital velocity.

We agree that the tangential velocity touches the orbit at only one place. I assume your orbital velocity vector would touch the ellipse at two places. Do I understand you correctly?

Does the magenta vector represent your orbital velocity? If not, please describe how I should change the magenta vector to better represent your orbital velocity. I don't need exact numbers, just descriptions such as:
make it longer or shorter
change the direction 10 degrees up or down.
Something like that. Also, remember the length is a geometric representation of speed. And also keep in mind that I have made all the vectors about 1000 times longer that they should be to make the illustration easier to see.
Image


Back to top
Profile WWW
Moderator
Moderator
avatar
Joined: Thu Jun 03, 2004 11:23 am
Posts: 3745
Location: Hamburg, Germany
Post    Posted on: Mon Jan 17, 2005 7:53 am
Please let me leave my answer to your post and the questions included to my next post because it seems to me that another clarification is required.

A tangent is a straight line touching a curved line at only one point - the three black vectors are lying upon such tangents. This seem to be a sufficient reason for you to consider these vectors to be "equal to" or "identical with" the velocity along the orbit, the ellipse or the circle.

You are right the tangenbtial vectors arte touching the ellipse, orbit, circle at only one point. This means mathematically that the slope of the tangent and the slope of the ellipse, circle, orbit are identical at that point - not that I'm speaking mathematically now.

To draw your orbits, ellipses, circles you are using a system of two coordinates whith the sun in the center where the two coordinates are intersecting each other. Your have drawn the tangents which the vectors are lying at into this coordinate system.

This means that the slope of the tangents and the vectors lying on them as well as the slope of the ellipses, circles and orbits always are slopes along one of the two coordiantes in relation to the remaining other coordinate.

From all this and the additional fact that none of the coordinates is representing time but both are representing directions wthin space... - from all this follows that the slope of the tangent cannot be representing velocity no way.

This in turn means necessaryly - logically - that in your diagrams and graphics the tangents and tangential vectors only show the direction of movement at the touching point which means that it is an instantanious direction - but NOT an instantaneous VELOCITY. And this is valid concerning YOUR GRAPHICS and DIAGRAMS - I'M NOT speaking about MY mathematics this moment, I don't look at the movements the way of such diagrams and graphics as I explained.

So from your diagrams etc. there is no way to conclude that the velocity along the tangents is identical to the velocity along the ellipse etc. at the touching point - your diagram don't include time!

The only way to include VELOCITY into your diagrams is to draw the vectors as long as they must be in relation to the distances represented by the scale of the coordinates. If 15 cm at the coordinates are representing 150,000,000 km a tangential velocity vector representing 30 km/s must be 3 * 15 / 15,000,000 cm = 3/1,000,000 = 3 * 10^-6 cm long! Then all would be within the scale and it would be possible to see a little bit what I#m doing and thinking.

A last point is left within these considerations - the instantaneous direction at the touching point is changed at the next following point. This is caused by the vertical vector pointing to the sun. To get the movement along the orbit this vector has to be included and it means that to get velocity the change of direction has to be included - so the instantaneous direction is out of consideration.

Please think abou it - the answer(s) to your questions will folow in my next post.



Dipl.-Volkswirt (bdvb) Augustin (Political Economist)

EDIT: There is another point meaning that the slope of the tangent cannot be the velocity at the touching point - in your diagrams there would be two points at one orbit at which velocity would be zero because slope is zero and two points where velocity would be infinite because slope is infinite.

So slope cannot mean velocity within your diagrams. I don't believe that you take slope as velocity but wanted to clarify this because you seem to use the term "instantaneous" for the point of touch between the tangent and the orbit. This term is invalid within the diagram because the diagram doesn't include time - the term can only be valid if you include time another way than by coordinates. This way isn't clear to me. Veocity can easyly be included by the length of the vector along the tangent and would be tangential velocity then - but not orbital velocity of the orbit touched.


Back to top
Profile
Moon Mission Member
Moon Mission Member
User avatar
Joined: Tue Oct 05, 2004 5:38 pm
Posts: 1361
Location: Austin, Texas
Post    Posted on: Tue Jan 18, 2005 12:32 am
I think I see your confusion now.

You are thinking of a graph of distance Vs time. For example the X-axis represents time and the Y-axis represents distance. Then the slope represents speed. There is no direction information in the graph so such a graph cannot show a velocity vector. It only shows a speed scalar.

The diagrams I have been posting show position, velocity and acceleration all at the same time. Time is not represented at all. Position is just a point in the X, Y plane. For example, aphelion may be at the point (X, Y)=(150,000,000, 0). Velocity is shown as a vector, or arrow, in the diagram. The length of the arrow is the speed and, naturally, the direction of the arrow is the direction of the velocity. For example a velocity of 20 km/s at aphelion would be an arrow starting at (X, Y) = (150,000,000, 0) and ending at (150,000,000, 20). EXACTLY the SAME velocity could be shown as an arrow 1,728,000 km long starting at (X, Y) = (150,000,000, 0) and ending at (150,000,000, 1,728,000). These two entirely different representations of velocity are equal because they both represent the same RATIO of distance over time. So 20 km / 1 s = 20 km/s AND 1,728,000 km / 1 day = 20 km/s because there are 86,400 seconds in a day. For velocity, it is only the RATIO which matters. You could think of the diagram as showing the velocity with an arrow of length equal to the DISTANCE that WOULD be traveled IF it continued in uniform motion for the specified time interval. The choice of time interval does not matter at all, as long as the SAME time interval is used for all the vectors.

Acceleration can also be represented. It would be shown as a velocity vector, the velocity that would result if the acceleration continued unchanged for the specified time interval. For example, an acceleration of 0.005 km/s^2 toward the left along the X-axis could be shown as an arrow starting at (X, Y)=(150,000,000, 0) and ending at (X, Y)=(149,999,999.995, 0). It could also be represented as an arrow 86,400 times longer if the time interval was 1 day instead of 1 second.


Back to top
Profile WWW
Moderator
Moderator
avatar
Joined: Thu Jun 03, 2004 11:23 am
Posts: 3745
Location: Hamburg, Germany
Post    Posted on: Tue Jan 18, 2005 8:37 am
Peter,

the confusion you diagnosting isn't at my side. No way.

What you say - "The diagrams I have been posting show position, velocity and acceleration all at the same time. Time is not represented at all." is just your repetition of what I said in the post you are responding to.

If the lengths of the vectors to be seen in your diagrams are the expression of the real velocity at the point they are beginning at... - then they are representing velocities of a hundred thousands or millions of km per second and not 20 km/s or 30 km/s: look at my calculation in the post you are responding to. The fraction of the vectors you drew is too large compared to the total scale of the diagrams or the ellipses to represent such small velocities if the longer axes of the ellipses or the total lengths of the coordinate axises are representing 150,000,000 km. Your vectors are wrong scaled.

I explicitly said that I am considering YOUR mathematics and diagrams and what you say means that you agree to the result of my considerations. To consider YOUR mathematics and diagrams doesn't say anything about my methods, mathematics and the like - it does NO WAY. I don't interpret your graphs as having to do anything with time - in contrary - its obvious that they don't. But from your posts it seems to me that your understanding of your own graphics is including the implicit interpretation of the point of touch between the tangent and the ellipse as a special time. The reason why it seems to me to be so is that you are calling the vector touching the ellipse at that point "instantaneous velocity". The fact that no time axis is included in your diagrams means that nothing in your diagrams allows for the interpretation of the touching points as "instantaneous". What your diagram does allow for is the interpretation of these points as "local". Perhaps you are not conscious of this.

If you have in mind "instantaneous" then allways additional external explicit explanations for each diagram are required.

I'm not using your coordinate system and I'm not calculating ellipses and so I never have in mind your diagrams. I'm too not using no coordinate system including a time axis and so I don't use graphs "of distance Vs time".

Before I started my calculations I said something about my coordinate system and you felt remembered to the times before Kopernikus which I explicitly said to be wrong. I explained the reasons and you understood it as your answers showed.

You are doing conclusions from my post you are respoding to concerning my own thinking and calculations - and that's invalid because such conclusions are logically impossible.

Please explain the lengths of your vectors. If they are representing velocity why they are that long then. Please explain the scale of your diagrams and ellipses too - are the large axises of the ellipses millions of km long? ...

Within your diagram you could substitue your ellipses by a chain of vectors if the vectors fit into the scale. You easyly will see that the lot of vectors valid per second will be a much closer approximation to your current ellipses that the fewer vectors valid per hour or per day. That's very important.



Now the promised answer to your previous post:

The magenta vector is SIMILAR to my orbital velocity - please note "SIMILAR". YOUR magenta vector is too long and by far not that approximation that my vector is - as far as I can conclude this from your diagram (remember my questions concerning scale and lengths above).

You say that you made all the vectors 1000 times longer - that's wrong: you made them 1,000,000 times longer if the large axis of your ellipses or the diameter of your circles are 150,000,000 km long.

Your information that the vectors are that longer isn't a sufficient answer to my questions above because the representation of velocity by these vectors requires that they fit into the scale of the diagrams which is not the case if you make them that longer. This way you are doing that error that caused me spirals when I calculate qute other things.



Dipl.-Volkswirt (bdvb) Augustin (Political Economist)


Back to top
Profile
Moon Mission Member
Moon Mission Member
User avatar
Joined: Tue Oct 05, 2004 5:38 pm
Posts: 1361
Location: Austin, Texas
Post    Posted on: Tue Jan 18, 2005 2:40 pm
Ekkehard Augustin wrote:
Time is not represented at all." is just your repetition of what I said in the post you are responding to.

OK, I’ll agree with that.

Ekkehard Augustin wrote:
The fraction of the vectors you drew is too large compared to the total scale of the diagrams

Remember that velocity is the RATIO of a length and a TIME interval. By choosing a large time interval you can correctly show a very small velocity as a very long arrow in the diagram. 12,096,000 kilometers per week is the SAME velocity as 20 kilometers per second so that a 12,096,000 km line can mean the same as a 20 km line in the diagram. As long as ALL vectors in a diagram are drawn using the SAME time interval, the graph is correct. Also, I can show velocity in kilometers per week in the diagram and calculate with kilometers per second. This is usually done in physics when the vectors would be too small to see if shown using the small time interval being used in the calculations. Extra text is not usually considered necessary, although for actual calculations the vectors are usually labeled better than I have been labeling them.

Ekkehard Augustin wrote:
your understanding of your own graphics is including the implicit interpretation of the point of touch between the tangent and the ellipse as a special time. The reason why it seems to me to be so is that you are calling the vector touching the ellipse at that point "instantaneous velocity".

Exactly! You understand too.

Ekkehard Augustin wrote:
The fact that no time axis is included in your diagrams means that nothing in your diagrams allows for the interpretation of the touching points as "instantaneous". What your diagram does allow for is the interpretation of these points as "local". Perhaps you are not conscious of this.

OK, maybe you don’t understand.

Ekkehard Augustin wrote:
If you have in mind "instantaneous" then allways additional external explicit explanations for each diagram are required.

In physics, these types of diagrams are always understood to show instantaneous velocities and accelerations at the point in time when the orbiting object is at the location where the vectors are drawn. If more than one point has vectors drawn then it is understood that more than one instant in time is being shown.

Ekkehard Augustin wrote:
I'm not using your coordinate system and I'm not calculating ellipses and so I never have in mind your diagrams. Before I started my calculations I said something about my coordinate system and you felt remembered to the times before Kopernikus which I explicitly said to be wrong. I explained the reasons and you understood it as your answers showed.

I really think you need to change coordinate systems. When I first wrote a Moon lander game I assumed a flat surface with the LM landing on it. Gravity was a constant force downward and the LM could have a velocity composed of horizontal vertical components. The goal of the game was to land with both velocity components as low as possible. By rotating the LM, it’s rockets could be used to change both horizontal and vertical velocities. Gravity would always tend to increase the vertical velocity in a downward direction. This was OK for simulating the final seconds of landing, but I wanted to start in orbit, land, take off and return to orbit. I knew that a high enough horizontal speed would result in an orbit. (EDIT) And the LM would no longer fall because of gravity, even with no rocket power. (END EDIT) I also knew that the horizontal velocity should change as the LM orbited up to the high point and back down to the low point in it’s elliptical orbit. Since gravity never had any component in the horizontal direction I was puzzled how to formulate the math to show the changing horizontal speed in an elliptical orbit. Eventually I succeeded in writing some VERY complicated math using trigonometry that did correctly simulate elliptical orbits. This situation seems very similar to what you are doing in your orbit calculations, except your math does not include enough terms to correctly simulate a real orbit. Eventually I decided to change coordinate systems to one centered on the massive body being orbited. Then I realized that my original coordinate system had been constantly rotating to keep the horizontal velocity horizontal and this was complicating the math almost beyond recognition. The coordinate change made everything clear and simplified the math to an almost childish level. If I wanted, I could transform to polar coordinates and use the radial velocity as the old vertical velocity and the angular velocity as the old horizontal velocity. It is all so easy!

Ekkehard Augustin wrote:
Please explain the lengths of your vectors. If they are representing velocity why they are that long then. Please explain the scale of your diagrams and ellipses too - are the large axises of the ellipses millions of km long? ...

12,096,000 kilometers per week is the SAME velocity as 20 kilometers per second, so a 12,096,000 km line can mean the same thing as a 20 km line in the diagram. As long as ALL the vectors in the diagram are drawn using the SAME time interval, then the absolute length in the diagram is not important. The important information is that a vector twice as long as another represents a speed twice as fast. The diagram can show vectors in km/week while the calculations are done in km/second with no problem at all.


Ekkehard Augustin wrote:
Within your diagram you could substitue your ellipses by a chain of vectors if the vectors fit into the scale. You easyly will see that the lot of vectors valid per second will be a much closer approximation to your current ellipses that the fewer vectors valid per hour or per day. That's very important.

Yes, I agree completely. Smaller vectors result in a better result. But I must show the vectors big so they can be clearly seen in the diagrams. The giant vectors drawn would never work in a calculation, but the small vectors used for calculation could never be seen in the diagram. You need to be thinking at a higher level of abstraction as you compare graphs and calculations.


Ekkehard Augustin wrote:
The magenta vector is SIMILAR to my orbital velocity - please note "SIMILAR". YOUR magenta vector is too long and by far not that approximation that my vector is - as far as I can conclude this from your diagram (remember my questions concerning scale and lengths above).

All I want to know is if the magenta vector is the right size in relation to the size of the associated black vector. See my several explanations of size above. And I agree that the difference between the black and magenta vectors will be much larger because of the large size of both vectors in the diagram, but we can imagine them much smaller and imagine how similar they would be without actually drawing them so small that a microscope would be needed to see them. Again, this is a higher level of abstraction in thinking, but it is necessary to keep from getting bogged down in details. And we are VERY bogged down in details in this discussion.

So, with all of the above in mind, in the magenta vector ABOUT right? In particular, if the vectors were drawn as small as you would like, would then length of the magenta vector always be nearly the same as that of the black vector?


Back to top
Profile WWW
Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 160 posts ] 
Go to page Previous  1, 2, 3, 4, 5, 6, 7, 8 ... 11  Next
 

Who is online 

Users browsing this forum: No registered users and 12 guests


© 2014 The International Space Fellowship, developed by Gabitasoft Interactive. All Rights Reserved.  Privacy Policy | Terms of Use