✓ Accepted answer
I think that the easiest solution would be a for loop. Here is a basic example:
for i = .1, 1, .1 do
image.ImageTransparency = i
end
You should not do what the answer above tells you. He says to make the decimal amounts smaller that you add to the Transparency to get it to go slower yet he has a wait() that will wait almost no time. The better way to slow a loop is a wait() with a number you like in it instead of smaller and smaller values. So altogether our code should look like this:
for i = .1, 1, .1 do
image.ImageTransparency = i
wait(.1)
end
In your question you subtracted from the Transparency. Transparency means how invisible something is. So if you are subtracting from it what is going to happen? It is going to become less invisible or in normal speech: more visible. That is why either you should add to Transparency or make it change in increasingly larger values. Anyway, Hope I helped and have a great day scripting!
If you have any additional questions or concerns please post them in the comments below.
2 comments
I wrote this early in the morning so please forgive any grammar and spelling errors you may find.
Yeah, your way is better. I didn't mean to say that if he wanted to make the loop go slower, he should change the "0.01". I meant to say that he could change it to something like "0.05" if he wanted it to go faster.