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		<title>Cycling 74  &#187;  Topic: slide object question &#8230;</title>
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		<pubDate>Wed, 19 Jun 2013 00:28:55 +0000</pubDate>
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					<guid>http://cycling74.com/forums/topic/slide-object-question/#post-60477</guid>
					<title><![CDATA[slide object question &#8230;]]></title>
					<link>http://cycling74.com/forums/topic/slide-object-question/#post-60477</link>
					<pubDate>Mon, 05 Dec 2011 21:49:36 +0000</pubDate>
					<dc:creator>Per</dc:creator>

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						<p>Can someone please help me to understand the smoothing process of the slide object.</p>
<p>As in the Max Data Tutorial 2: Data Scaling it gives the example:<br />
For example, when you first start the patch, the “last” value is 0.0. If you make an immediate transition to 0.5 (which you can do by entering a value into the number box connected to the slide objects), you will see that the first slider moves to 0.01.</p>
<p>the difference between the initial value of 0.00 and 0.50 is 0.50; this difference is divided by the factor (in this case 50.) to yield 0.01.</p>
<p>that would cause the calculation:<br />
0.0 to 0.5 are equal to 0.5 in difference<br />
divided by factor 50 (0.5/50) with the result 0.01</p>
<p>so far so good. </p>
<p>but say the next incoming value would be 1.0<br />
that would then be:<br />
the difference between the value of 0.01 and 1.00 is 0.99<br />
and then divide it by 50 would cause 0.99/50=0.0198</p>
<p>but the result if Im using this example in Max Data Tutorial 2: Data Scaling patch is: 0.030</p>
<p>why??</p>
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					<guid>http://cycling74.com/forums/topic/slide-object-question/#post-217674</guid>
					<title><![CDATA[Re: slide object question &#8230;]]></title>
					<link>http://cycling74.com/forums/topic/slide-object-question/#post-217674</link>
					<pubDate>Tue, 06 Dec 2011 00:54:42 +0000</pubDate>
					<dc:creator>Roman Thilenius</dc:creator>

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						<p>because the averging isnt restarted when you change the argument while it runs? :)</p>
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					<guid>http://cycling74.com/forums/topic/slide-object-question/#post-217675</guid>
					<title><![CDATA[Re: slide object question &#8230;]]></title>
					<link>http://cycling74.com/forums/topic/slide-object-question/#post-217675</link>
					<pubDate>Wed, 07 Dec 2011 15:23:31 +0000</pubDate>
					<dc:creator>Per</dc:creator>

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						<p>Sorry but i didnt understand what you mean &#8230;</p>
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					<guid>http://cycling74.com/forums/topic/slide-object-question/#post-217676</guid>
					<title><![CDATA[Re: slide object question &#8230;]]></title>
					<link>http://cycling74.com/forums/topic/slide-object-question/#post-217676</link>
					<pubDate>Sun, 08 Jan 2012 13:23:23 +0000</pubDate>
					<dc:creator>Per</dc:creator>

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						<p>anyone who can help me explaining this?</p>
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					<guid>http://cycling74.com/forums/topic/slide-object-question/#post-217677</guid>
					<title><![CDATA[Re: slide object question &#8230;]]></title>
					<link>http://cycling74.com/forums/topic/slide-object-question/#post-217677</link>
					<pubDate>Mon, 09 Jan 2012 09:34:32 +0000</pubDate>
					<dc:creator>Patrick Delges</dc:creator>

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						<p>I hope you didn&#8217;t spend one month on this!</p>
<p>You can find <strong>slide</strong>&#8216;s formula in the reference:</p>
<p>y (n) = y (n-1) + ((x (n) &#8211; y (n-1))/slide)</p>
<p>So, if you use your example, y(2) = <strong>0.01 +</strong> (1 &#8211; 0.01) / 50 = 0.01 + 0.0198 = 0.0298</p>
<p>You forgot to add the previous value (y(n-1)) to your result (in this type of filter, a value always depends on the previous one).</p>
<p>Now, you may think that 0.03 is not 0. 0298, but if you make the float box bigger, you&#8217;ll discover that Max in fact also computes a 0.0298</p>
<p>HTH,</p>
<p>p</p>
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					<guid>http://cycling74.com/forums/topic/slide-object-question/#post-217678</guid>
					<title><![CDATA[Re: slide object question &#8230;]]></title>
					<link>http://cycling74.com/forums/topic/slide-object-question/#post-217678</link>
					<pubDate>Mon, 09 Jan 2012 15:06:25 +0000</pubDate>
					<dc:creator>Roman Thilenius</dc:creator>

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						<p>otherwise, put it on a katapult and &#8230;. wooosh.</p>
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