
zoom question
Hi,
Say you have a virtual map and its at 1x zoom (normal). The map is going to zoom to a scale of 10 in 7 seconds.
The time for transitions between zoom levels is not evenly distributed, since zooming from level 1 to 2 is a scale of 2, and zooming from 5 to 6 is a scale of 1.2.
I have control of the zoom of the map, but i can only zoom from level 1 to 3, 3 to 5, 5 to 8 then 8 to 10 in those increments. How do i calculate the time for each of these increments so the zoom appears completely smooth over the total time?
Just used example figures so if you can help please dont just give answer. Must be very simple for u mathematicians but cant get my head around combining scales distance and time. Many thanks for any help.

1,2,3,4,5,6,7,8,9,10.
the time from 1 to 10 is 7 seconds. We need to find the ammount of time nessicary for 1 transition to seem smooth.
$\displaystyle 10(x)=7$
$\displaystyle x=\frac{7}{10}=.7$ seconds for smoothness per zoom.
So, to figure out the smoothness time for a set of numbers we must find the difference (call the number 'r')
so
$\displaystyle \Delta(r)x=t$
so, 1 to 3.
$\displaystyle \Delta(r)=2 \therefore \,\,\,\,t=2(.7)=1.4s$
This is assuming a transition from 1 to 10 in 7 seconds is smooth.

Thanks for the help. Im not familiar with some of the symbols you have used but I will do some research.

$\displaystyle \Delta$ just means change in.
Say we have $\displaystyle \Delta(x)$ such that $\displaystyle x$ is defined by $\displaystyle n$ and $\displaystyle j$ so that $\displaystyle j>n$
Then by definition $\displaystyle \Delta(x)=jn$
In your case, we defined $\displaystyle \Delta(r)$ with respect to any two of your numbers. You get the two numbers and see how they satisfy $\displaystyle j>n$ then subtract. If they are equal, subtract to gain zero; meaning no change.
example: 1 to 3. Any set of two integer numbers over the interval of [1,10] can be represented as 'r' and thus we can say
$\displaystyle \Delta(r)=jn, j>n$
thus, with respect to 1 and 3
$\displaystyle \Delta(r)=31, \because 3>1$ where the upside down triangle is a symbol for because.

Hi again thanks for your help.
I know how to distribute the time evenly, my situation is more complicated than this.
If the transition between each zoom level takes the same 0.7 seconds then the effect will not appear smooth. It will appear to slow down near the end. Imagine a picture in front of you, it moves towards you, in 0.7 seconds you are at zoom level 2 and can see 2X the level of detail you did before.
In 1.4 seconds (level 3) you can see only 1.5X the level of detail you did at 0.7 seconds. The increase in detail seen is not mapped evenly over the total time.
Going from 1 to 10, no pauses between transitions.
1  3: 3X Zoom
3  5: 1.66X Zoom
5  8: 1.6X Zoom
8  10 : 1.25X Zoom
I made a quick swf to show exactly what the problem is:
View SWF File

Oh, well, I don't know (Nerd)

Ive done some research and it appears to be something to do with logarithmic scale. If anyone can show me how to apply this it would be really useful thanks.

bump  can anyone help with logarithmic scale and time?

Hi bumping. If im going to apply logarithmic time scaling I will have to distribute the time for each and every level if necessary I think. Can anyone please show me how this is done?