Hi All,
mxc_v4l2_capture.c file in below content what basis they hard coded .
cam->v2f.fmt.pix.sizeimage = 352 * 288 * 3 / 2; // expalin me this values what basis
cam->v2f.fmt.pix.bytesperline = 288 * 3 / 2;
cam->v2f.fmt.pix.width = 288;
cam->v2f.fmt.pix.height = 352;
Crop bound width and height is 640*480 but how v2f format width is 288 how ?
Would you please help me any one , I am beginner of the camera driver .
Thanks
Viswa
Hi Viswa
these calculations are for memory requirements, how many bytes
needed for one pixel. Also please check Table 37-2. Data Formats Supported By The
Camera Port i.MX6DQ Reference Manual
http://cache.freescale.com/files/32bit/doc/ref_manual/IMX6DQRM.pdf
and AN5305 MIPI–CSI2 Peripheral on i.MX6 MPUs
http://www.nxp.com/assets/documents/data/en/application-notes/AN5305.pdf
Best regards
igor
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Hi Igor,
Thanks for your quick reply.
i verified the imx6DQRM.pd .
This for formula :
F = FH * FW * FPS * BI * DF
crop bound height 640
crop bound width 480
sorry I am not understanding,
if you possible, Would you please explain me in detail what basis they calculated
Thanks
Viswa