Return-Path: Received: from mail-vx0-f169.google.com ([209.85.220.169] verified) by media-motion.tv (CommuniGate Pro SMTP 4.2.10) with ESMTP-TLS id 4649460 for AE-List@media-motion.tv; Thu, 08 Mar 2012 18:27:40 +0100 Received: by vcbfk14 with SMTP id fk14so605481vcb.28 for ; Thu, 08 Mar 2012 09:34:37 -0800 (PST) DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=gmail.com; s=20120113; h=mime-version:in-reply-to:references:date:message-id:subject:from:to :content-type; bh=L1g0tjjL1uyTBjNfcMD2E/pyUQEUZrj5e2r4YqYxE28=; b=ja5Ohu7OtPbV3MpAgAcp06kV9qxlt/rY8O8ALymd4whrA/4xWhkH3f7OZcBqn6ZvIr Q+dFDN5WjBEP2oAy2cnZivLB1dUiPb+LsgHVc0qN3QA49ymb69EIGck+gl1bZuTxC9dL lHSFwQuJyIxiCHrVPpf06dRlPmdcfkwj+FqN3zytewM5iEKxHCRaW8lcYR5+Jzmmw5+x edbhmEGikA0iqFrU0ouAkaNi6UhntuDYqr5x8iMLel5/H2O8dl4NQRC2GqZfxImjEDrQ 9bgrdsPF8JngT0OFlLvN90zRX9WURWqZU/o86XP/ZNNXdXmHpLaI3Z9qa5rOh3pHG6dN sSUQ== MIME-Version: 1.0 Received: by 10.52.94.148 with SMTP id dc20mr11257111vdb.100.1331228077315; Thu, 08 Mar 2012 09:34:37 -0800 (PST) Received: by 10.220.32.207 with HTTP; Thu, 8 Mar 2012 09:34:37 -0800 (PST) In-Reply-To: References: Date: Thu, 8 Mar 2012 09:34:37 -0800 Message-ID: Subject: Re: [AE] Expression while loops From: Darby Edelen To: After Effects Mail List Content-Type: multipart/alternative; boundary=bcaec5015f2ba9d72604babeb073 --bcaec5015f2ba9d72604babeb073 Content-Type: text/plain; charset=windows-1252 Content-Transfer-Encoding: quoted-printable As for Math.exp(x) that will depend on how you're using it. Usually as x increases the result will increase exponentially (faster) as you've described. If that's not the behavior then there may be some specifics in the way you're using it that are causing the result to behave differently. Possible variations might include: Math.exp(x) =96 returns 1 when x is 0 and exponentially larger values when = x is above 0 Math.exp(-x) or 1 / Math.exp(x) =96 returns 1 when x is 0 and exponentially smaller values when x is above 0 Math.exp(1/x) =96 undefined at 0 (divide by 0, a vertical asymptote) but returns values ranging from infinity to e as x goes from 0 to 1. Approaches horizontal asymptote of 1 as x increases from 1 to infinity. -D On Wed, Mar 7, 2012 at 10:08 AM, Gary Berendsen wro= te: > > Also another question how do you invert the exponential curve from > math.exp so things start slower and go fster at the end? > > -- > gary berendsen > http://garyberendsen.com > VFX generalist > > > +---End of message---+ > To unsubscribe send any message to > --bcaec5015f2ba9d72604babeb073 Content-Type: text/html; charset=windows-1252 Content-Transfer-Encoding: quoted-printable As for Math.exp(x) that will depend on how you're using it.=A0 Usually = as x increases the result will increase exponentially (faster) as you'v= e described.=A0 If that's not the behavior then there may be some speci= fics in the way you're using it that are causing the result to behave d= ifferently.

Possible variations might include:

Math.exp(x) =96 returns 1 whe= n x is 0 and exponentially larger values when x is above 0

Math.exp(= -x) or 1 / Math.exp(x) =96 returns 1 when x is 0 and exponentially smaller = values when x is above 0

Math.exp(1/x) =96 undefined at 0 (divide by 0, a vertical asymptote) bu= t returns values ranging from infinity to e as x goes from 0 to 1.=A0 Appro= aches horizontal asymptote of 1 as x increases from 1 to infinity.

-= D

On Wed, Mar 7, 2012 at 10:08 AM, Gary Berend= sen <gary@ga= ryberendsen.com> wrote:

Also another question how do you invert the exponential curve from math.exp= so things start slower and go fster at the end?

--
gary berendsen
http://garyberendsen= .com
VFX generalist


+---End of message---+
To unsubscribe send any message to <ae-list-off@media-motion.tv>

--bcaec5015f2ba9d72604babeb073--