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where D is defined by Eq. (27.2), and the three coefficients are obtained from Table 27.4. Shaft Deviations. For shafts designated a through h, the upper deviation is equal to the fundamental deviation. Subtract the IT grade from the fundamental deviation to get the lower deviation. Remember, the deviations are defined as algebraic, so be careful with signs. Shafts designated j through zc have the lower deviation equal to the fundamental deviation. For these, the upper deviation is the sum of the IT grade and the fundamental deviation. Hole Deviations. Holes designated A through H have a lower deviation equal to the negative of the upper deviation for shafts. Holes designated as J through ZC have an upper deviation equal to the negative of the lower deviation for shafts. An exception to the rule occurs for a hole designated as N having an IT grade from 9 to 16 inclusive and a size over 3 mm. For these, the fundamental deviation is zero. A second exception occurs for holes J, K, M, and N up to grade IT8 inclusive and holes P through ZC up to grade 7 inclusive for sizes over 3 mm. For these, the upper deviation of the hole is equal to the negative of the lower deviation of the shaft plus the change in tolerance of that grade and the next finer grade. In equation form, this can be written Upper deviation (hole) = lower deviation (shaft) + IT (shaft) IT (next finer shaft) (27.4)

Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright 2004 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.

Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright 2004 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.

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-(void)createOffscreenCtx { NSOperationQueue *q = [(BigViewThingAppDelegate *) [[UIApplication sharedApplication] delegate] globalQ]; PLBlockOperation *op = [PLBlockOperation blockOperationWithBlock:^{ //imgRef = [[UIImage imageNamed:imageName] CGImage]; NSString* bundlePath = [[NSBundle mainBundle] bundlePath]; UIImage *img = [UIImage imageWithContentsOfFile:[NSString stringWithFormat:@"%@/%@", bundlePath, imageName]]; CGImageRef imgRef = [img CGImage]; CGSize layerSize = [self bounds].size; layerSize.height = floorf(layerSize.height); layerSize.width = floorf(layerSize.width); CGColorSpaceRef colorSpace = CGColorSpaceCreateDeviceRGB(); CGContextRef ctx = (CGContextRef) [(id) CGBitmapContextCreate(NULL, layerSize.width, layerSize.height, 8, layerSize.width*4, colorSpace, kCGImageAlphaPremultipliedLast) autorelease]; CGColorSpaceRelease(colorSpace); CGContextTranslateCTM(ctx, 0, layerSize.height); CGContextScaleCTM(ctx, 1.0, -1.0); CGFloat tx = layerSize.width * (1.0 - scale) * 0.5; CGFloat ty = layerSize.height * (1.0 - scale) * 0.5;

Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright 2004 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.

Up to 32 devices can be connected to an RS-485 data line with the SN75176 and communicate by using a data protocol This is referred to as an RS-485 network, or an RS-485 drop network The heart of this communication is the Texas Instruments SN75176 differential bus transceiver chip This chip converts RS-485 signals to RS-232 TTL-level signals, allowing devices that traditionally communicate over standard RS-232 serial connections to communicate over a single two-wire RS-485 network The Texas Instruments SN75176 chip has only 8 pins, and it can be easily connected to the BASIC STAMP, as shown in Fig 2-16 The SN75176 requires 30 mA; therefore, it is recommended that only one of these chips be driven from the STAMP s Vdd pin, provided no other components require current.

27.2.4 Preferred Fits Table 27.5 lists the preferred fits for most common applications. Either first or second choices from Table 27.3 should be used for the basic sizes. Example 1. Using the shaft-basis system, find the limits for both members using a basic size of 25 mm and a free-running fit. Solution. From Table 27.5, we find the fit symbol as D9/h9, the same as Fig. 27.2. Table 27.1 gives Dmin = 18 and Dmax = 30 for a basic size of 25. Using Eq. (27.2), we find D= DmaxDmin = 30 (18) = 23.2 mm

Then, from Eq. (27.1) and Table 27.2, 40i = = 40 (0.45D1/3 + 0.001D) 1000 40 [0.45(23.2)1/3 + 0.001(23.2)] = 0.052 mm 1000

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