{"id":29106,"date":"2022-02-10T00:00:00","date_gmt":"2022-02-10T00:00:00","guid":{"rendered":"https:\/\/c01.purpledshub.com\/bbcskyatnight\/?post_type=purple_issue&#038;p=29106"},"modified":"2022-03-23T13:16:36","modified_gmt":"2022-03-23T13:16:36","slug":"diy-astronomy-5","status":"publish","type":"post","link":"https:\/\/c01.purpledshub.com\/bbcskyatnight\/2022\/02\/10\/diy-astronomy-5\/","title":{"rendered":"DIY Astronomy"},"content":{"rendered":"\n<h4 class=\"has-text-align-center article-full-subhead\">Use Moon shadows to measure crater peaks<\/h4>\n\n<p class=\"has-text-align-center intro\">How a simple formula gives you a deeper understanding of the physical form of lunar craters<\/p>\n\n<div class=\"no-tts wp-block-image\"><figure class=\"no-tts aligncenter size-large\"><img loading=\"lazy\" width=\"1024\" height=\"953\" src=\"https:\/\/dj9jqhxgw9833.cloudfront.net\/uploads\/sites\/77\/2022\/02\/2L2E65QN67X664N973M04Q22RE5D-1024x953.jpg\" alt=\"\" class=\"no-tts wp-image-29328\" srcset=\"https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/2L2E65QN67X664N973M04Q22RE5D-1024x953.jpg 1024w, https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/2L2E65QN67X664N973M04Q22RE5D-300x279.jpg 300w, https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/2L2E65QN67X664N973M04Q22RE5D-768x715.jpg 768w, https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/2L2E65QN67X664N973M04Q22RE5D-1536x1430.jpg 1536w, https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/2L2E65QN67X664N973M04Q22RE5D.jpg 1858w\" sizes=\"(max-width: 1024px) 100vw, 1024px\" \/><figcaption>Lunar craters like Theophilus shown here, with clear wall shadows and an interesting central feature, make good examples for measuring physical heights<\/figcaption><\/figure><\/div>\n\n<div class=\"no-tts wp-block-image\"><figure class=\"no-tts aligncenter size-large\"><img loading=\"lazy\" width=\"1024\" height=\"683\" src=\"https:\/\/dj9jqhxgw9833.cloudfront.net\/uploads\/sites\/77\/2022\/02\/VH8531ENFRX6D183FQ35ROEZK35T-1024x683.jpg\" alt=\"\" class=\"no-tts wp-image-29332\" srcset=\"https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/VH8531ENFRX6D183FQ35ROEZK35T-1024x683.jpg 1024w, https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/VH8531ENFRX6D183FQ35ROEZK35T-300x200.jpg 300w, https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/VH8531ENFRX6D183FQ35ROEZK35T-768x512.jpg 768w, https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/VH8531ENFRX6D183FQ35ROEZK35T-1536x1024.jpg 1536w, https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/VH8531ENFRX6D183FQ35ROEZK35T.jpg 2048w\" sizes=\"(max-width: 1024px) 100vw, 1024px\" \/><figcaption>A graph comparing our measurements for the crater Theophilus with the published data shows the results are similar<\/figcaption><\/figure><\/div>\n\n<p class=\"has-drop-cap article-full-body sans-serif\">The lunar surface is at its most striking when there are well-defined shadows. As well as being visually pleasing, the study of shadows cast by lunar mountains and craters can tell us a lot about these features. When we see a lunar crater from above it\u2019s easy to think that the crater wall is uniform around the entire perimeter. However, if we look at the shape of the shadows cast by a crater wall, we can get a clearer sense of the peaks and troughs along the rim; and if the crater has a central peak its shadows will reveal its nature too.<\/p>\n\n<p class=\"article-full-body sans-serif\">Here, we will show you a straightforward method to measure these shadows. By using some simple rightangled triangle trigonometry we can calculate the height of the lunar feature casting the shadow. Our equation is: O = tan\u03b8 x A; where O = \u2018opposite\u2019 (ie the height of the feature), tan\u03b8 = the tangent of the Sun angle, and A = \u2018adjacent\u2019 (ie the length of the shadow).<\/p>\n\n<h5 class=\"article-subhead\"><strong>Timing is important<\/strong><\/h5>\n\n<p class=\"article-full-body sans-serif\">To get started all you need is a ruler, a calculator and a photo of a crater that has clear shadows. You need the time and date the photo was taken, so you can use the Lunar Terminator Visualization Tool (<a href=\"http:\/\/bit.ly\/3taEmyf\" data-type=\"URL\" data-id=\"bit.ly\/3taEmyf\">bit.ly\/3taEmyf<\/a>) to find out what the Sun angle was at that time, at any point on the lunar surface. We used a photo of crater Theophilus taken (by Alessandro<span> Bianconi) at 03:48 UT on 18 September 2011.<\/span><\/p>\n\n<p class=\"article-full-body sans-serif\">It is important to mention that our process here has been simplified. This method relies on a published crater diameter so we can scale up our shadow measurement. Most craters are not perfectly circular, so the published figure is an average; we only took one diameter measurement. If you use this calculation for an isolated feature, you\u2019ll need to know the pixel\/ kilometre ratio for the equipment used to take photo.<\/p>\n\n<p class=\"article-full-body sans-serif\">Additionally, our process doesn\u2019t take the curvature of the lunar surface into account. Remember, the Moon is a sphere, so if you chose a crater that is quite central, the foreshortening effects are less apparent.<span> It can also be difficult to know exactly where the shadow starts and ends if it\u2019s located in a complex region. We are only using a single measurement of a complex crater at one Sun angle and comparing that to a published figure, which will be an average value.<\/span><\/p>\n\n<p class=\"article-full-body sans-serif\">Our method brings a sense of scale to an otherwise abstract landscape. Even though it has been simplified to make the maths easier, it still yields results that are close to the published figure. This project will help you to gain a deeper awareness of the lunar features you choose to analyse.<\/p>\n\n<section class=\"wp-block-uagb-section uagb-section__wrap uagb-section__background-color uagb-block-a7084eeb-53ad-4e95-b085-8b152bc38ed9 article-boxout\"><div class=\"uagb-section__overlay\"><\/div><div class=\"uagb-section__inner-wrap\">\n<h4 class=\"article-subhead\"><strong>What you\u2019ll need<\/strong><\/h4>\n\n\n\n<p class=\"article-full-body sans-serif\">A high-resolution digital photo of a lunar crater that has clearly defined shadows; we used a photo of the crater Theophilus.<\/p>\n\n\n\n<p class=\"article-full-body sans-serif\">A ruler; we used it to measure the shadow lengths on the computer screen, but you can use Photoshop\u2019s measuring tool or a printed photo.<\/p>\n\n\n\n<p class=\"article-full-body sans-serif\">A scientific calculator with a tan button. If you use Excel for these calculations, convert the Sun angle from degrees to radians first.<\/p>\n\n\n\n<p class=\"article-full-body sans-serif\">A copy of the Lunar Terminator Visualization Tool (VLT), which you can download from <strong><a href=\"http:\/\/bit.ly\/3taEmyf\">bit.ly\/3taEmyf<\/a>. <\/strong><\/p>\n\n\n\n<p class=\"article-full-body sans-serif\">The published crater diameter figure for the crater you\u2019re measuring from this online list of near-side craters: <strong><a href=\"http:\/\/bit.ly\/3GQNFYb\">bit.ly\/3GQNFYb<\/a> <\/strong><\/p>\n<\/div><\/section>\n\n<h4 class=\"has-text-align-center article-subhead\"><strong>Step by step<\/strong><\/h4>\n\n<h5 class=\"has-text-align-center article-subhead\"><strong>Step 1<\/strong><\/h5>\n\n<div class=\"no-tts wp-block-image article-in-image bild\"><figure class=\"no-tts aligncenter size-large is-resized\"><img loading=\"lazy\" src=\"https:\/\/dj9jqhxgw9833.cloudfront.net\/uploads\/sites\/77\/2022\/02\/5ZBT05G6UHROO98W3M9V6VLBG8VQ-1024x943.jpg\" alt=\"\" class=\"no-tts wp-image-29333\" width=\"512\" height=\"472\" srcset=\"https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/5ZBT05G6UHROO98W3M9V6VLBG8VQ-1024x943.jpg 1024w, https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/5ZBT05G6UHROO98W3M9V6VLBG8VQ-300x276.jpg 300w, https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/5ZBT05G6UHROO98W3M9V6VLBG8VQ-768x707.jpg 768w, https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/5ZBT05G6UHROO98W3M9V6VLBG8VQ-1536x1414.jpg 1536w, https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/5ZBT05G6UHROO98W3M9V6VLBG8VQ.jpg 1702w\" sizes=\"(max-width: 512px) 100vw, 512px\" \/><\/figure><\/div>\n\n<p class=\"has-text-align-center article-full-body sans-serif\">Open your crater photo and view it full screen. Next, use a ruler to measure the diameter of the crater. We measured our example, Theophilus, at the widest point in the same direction that the shadows lie and got a figure of 11.5cm.<\/p>\n\n<h5 class=\"has-text-align-center article-subhead\"><strong>Step 2<\/strong><\/h5>\n\n<div class=\"no-tts wp-block-image article-in-image photo\"><figure class=\"no-tts aligncenter size-large is-resized\"><img loading=\"lazy\" src=\"https:\/\/dj9jqhxgw9833.cloudfront.net\/uploads\/sites\/77\/2022\/02\/3ED2329732J0C6W2L17RQNG6PI18-1024x926.jpg\" alt=\"\" class=\"no-tts wp-image-29334\" width=\"512\" height=\"463\" srcset=\"https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/3ED2329732J0C6W2L17RQNG6PI18-1024x926.jpg 1024w, https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/3ED2329732J0C6W2L17RQNG6PI18-300x271.jpg 300w, https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/3ED2329732J0C6W2L17RQNG6PI18-768x695.jpg 768w, https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/3ED2329732J0C6W2L17RQNG6PI18-1536x1390.jpg 1536w, https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/3ED2329732J0C6W2L17RQNG6PI18.jpg 2003w\" sizes=\"(max-width: 512px) 100vw, 512px\" \/><figcaption>Calculation of the real shadow length: (110,000m \u00f7 11.5cm) x 3.5cm = 33,478m<\/figcaption><\/figure><\/div>\n\n<p class=\"has-text-align-center article-full-body sans-serif\">Measure the crater wall shadow; we got a value of 3.5cm. Calculate the real length of the shadow by taking the published diameter (110,000m) and dividing by the diameter on screen (11.5cm), then multiply by the shadow length on screen (3.5cm) to get 33,478m.<\/p>\n\n<h5 class=\"has-text-align-center article-subhead\"><strong>Step 3<\/strong><\/h5>\n\n<div class=\"no-tts wp-block-image article-in-image photo\"><figure class=\"no-tts aligncenter size-large is-resized\"><img loading=\"lazy\" src=\"https:\/\/dj9jqhxgw9833.cloudfront.net\/uploads\/sites\/77\/2022\/02\/2C540J8P2WMRZYQY9G3MESDZ7ZO5-1024x964.jpg\" alt=\"\" class=\"no-tts wp-image-29335\" width=\"512\" height=\"482\" srcset=\"https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/2C540J8P2WMRZYQY9G3MESDZ7ZO5-1024x964.jpg 1024w, https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/2C540J8P2WMRZYQY9G3MESDZ7ZO5-300x283.jpg 300w, https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/2C540J8P2WMRZYQY9G3MESDZ7ZO5-768x723.jpg 768w, https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/2C540J8P2WMRZYQY9G3MESDZ7ZO5-1536x1447.jpg 1536w, https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/2C540J8P2WMRZYQY9G3MESDZ7ZO5.jpg 1717w\" sizes=\"(max-width: 512px) 100vw, 512px\" \/><figcaption>Calculation of the real shadow length: (110,000m \u00f7 11.5cm) x 2.3cm = 22,000m<\/figcaption><\/figure><\/div>\n\n<p class=\"has-text-align-center article-full-body sans-serif\">Measure the crater\u2019s central peak shadow; we got a value of 2.3cm. Calculate the real length of the shadow by taking the published diameter (110,000m) and divide by the one on screen (11.5cm), then multiply by the shadow length on screen (2.3cm) to get 22,000m.<\/p>\n\n<h5 class=\"has-text-align-center article-subhead\"><strong>Step 4<\/strong><\/h5>\n\n<div class=\"no-tts wp-block-image article-in-image bild\"><figure class=\"no-tts aligncenter size-large is-resized\"><img loading=\"lazy\" src=\"https:\/\/dj9jqhxgw9833.cloudfront.net\/uploads\/sites\/77\/2022\/02\/7G9D554ADUS3P20052ZE2W9J7Y6R-1024x762.jpg\" alt=\"\" class=\"no-tts wp-image-29336\" width=\"512\" height=\"381\" srcset=\"https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/7G9D554ADUS3P20052ZE2W9J7Y6R-1024x762.jpg 1024w, https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/7G9D554ADUS3P20052ZE2W9J7Y6R-300x223.jpg 300w, https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/7G9D554ADUS3P20052ZE2W9J7Y6R-768x571.jpg 768w, https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/7G9D554ADUS3P20052ZE2W9J7Y6R-1536x1143.jpg 1536w, https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/7G9D554ADUS3P20052ZE2W9J7Y6R.jpg 1563w\" sizes=\"(max-width: 512px) 100vw, 512px\" \/><\/figure><\/div>\n\n<p class=\"has-text-align-center article-full-body sans-serif\">Open the Lunar Terminator Visualization Tool (<a href=\"http:\/\/bit.ly\/3taEmyf\">bit.ly\/3taEmyf<\/a>) and input the observation\u2019s date and time. Point the mouse at the point you began the crater wall shadow measurement and read the Sun angle; we got 4.87\u00b0. Repeat for the central peak; we got 3.67\u00b0.<\/p>\n\n<h5 class=\"has-text-align-center article-subhead\"><strong>Step 5<\/strong><\/h5>\n\n<div class=\"no-tts wp-block-image article-in-image bild\"><figure class=\"no-tts aligncenter size-large is-resized\"><img loading=\"lazy\" src=\"https:\/\/dj9jqhxgw9833.cloudfront.net\/uploads\/sites\/77\/2022\/02\/5GZ7QF12B5AMAXMKRJ6HBLKO9484.jpg\" alt=\"\" class=\"no-tts wp-image-29337\" width=\"499\" height=\"291\" srcset=\"https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/5GZ7QF12B5AMAXMKRJ6HBLKO9484.jpg 997w, https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/5GZ7QF12B5AMAXMKRJ6HBLKO9484-300x175.jpg 300w, https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/5GZ7QF12B5AMAXMKRJ6HBLKO9484-768x448.jpg 768w\" sizes=\"(max-width: 499px) 100vw, 499px\" \/><\/figure><\/div>\n\n<p class=\"has-text-align-center article-full-body sans-serif\">To calculate the height of the crater wall, use O = tan\u03b8 x A, where tan\u03b8 = tan4.87\u00b0 and A = shadow length of 33,478m. When we applied this to the crater wall in our example, we calculated a height of 2,852m compared to the published value of 3,200m.<\/p>\n\n<h5 class=\"has-text-align-center article-subhead\"><strong>Step 6<\/strong><\/h5>\n\n<div class=\"no-tts wp-block-image article-in-image bild\"><figure class=\"no-tts aligncenter size-large is-resized\"><img loading=\"lazy\" src=\"https:\/\/dj9jqhxgw9833.cloudfront.net\/uploads\/sites\/77\/2022\/02\/U73C837096C498HQ5X27KJ77GW02.jpg\" alt=\"\" class=\"no-tts wp-image-29338\" width=\"499\" height=\"291\" srcset=\"https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/U73C837096C498HQ5X27KJ77GW02.jpg 997w, https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/U73C837096C498HQ5X27KJ77GW02-300x175.jpg 300w, https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/U73C837096C498HQ5X27KJ77GW02-768x448.jpg 768w\" sizes=\"(max-width: 499px) 100vw, 499px\" \/><\/figure><\/div>\n\n<p class=\"has-text-align-center article-full-body sans-serif\">To calculate the height of the crater\u2019s central peak, use O = tan\u03b8 x A, where tan\u03b8 = tan3.67\u00b0 and A = shadow length of 22,000m. In our example, we calculated a height of 1,411m, compared to the published value of 1,400m.<\/p>\n\n<hr class=\"no-tts wp-block-separator is-style-wide\"\/>\n\n<div class=\"wp-block-columns bio\">\n<div class=\"wp-block-column bio_left\" style=\"flex-basis:33.33%\">\n<div class=\"no-tts wp-block-image\"><figure class=\"no-tts alignright size-large is-resized\"><img loading=\"lazy\" src=\"https:\/\/dj9jqhxgw9833.cloudfront.net\/uploads\/sites\/77\/2022\/02\/EU50886TR7OLPO671CO5A7W636PN.jpg\" alt=\"\" class=\"no-tts wp-image-29340\" width=\"136\" height=\"136\" srcset=\"https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/EU50886TR7OLPO671CO5A7W636PN.jpg 990w, https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/EU50886TR7OLPO671CO5A7W636PN-300x300.jpg 300w, https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/EU50886TR7OLPO671CO5A7W636PN-150x150.jpg 150w, https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/EU50886TR7OLPO671CO5A7W636PN-768x768.jpg 768w\" sizes=\"(max-width: 136px) 100vw, 136px\" \/><\/figure><\/div>\n<\/div>\n\n\n\n<div class=\"wp-block-column is-vertically-aligned-center bio_right\" style=\"flex-basis:66.66%\">\n<p>Mary McIntyre is an outreach astronomer and teacher of astrophotography<\/p>\n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>How a simple formula gives you a deeper understanding of the physical form of lunar craters<\/p>\n","protected":false},"author":24,"featured_media":29097,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"ub_ctt_via":"","purple_page_number":"74","purple_custom_meta_purple_page_number":"74","purple_seq_number":"1","purple_custom_meta_purple_seq_number":"1","purple_source_article":"article_74-1.xml","purple_custom_meta_purple_source_article":"article_74-1.xml","purple_source_issue":"March-2022","purple_custom_meta_purple_source_issue":"March-2022","purple_external_id":"March-2022-74-1","purple_custom_meta_purple_external_id":"March-2022-74-1","purple_issue_code":"|0000086550||","purple_custom_meta_purple_issue_code":"|0000086550||","purple_android_product":"com.im.skyatnight.202","purple_custom_meta_purple_android_product":"com.im.skyatnight.202","purple_ios_product":"com.im.skyatnight.202","purple_custom_meta_purple_ios_product":"com.im.skyatnight.202","purple_web_product":"","purple_custom_meta_purple_web_product":"","purple_publication_id":"075fab74-0a21-4201-866a-899d6c41c40c","purple_migrated":"","kt_blocks_editor_width":""},"categories":[21],"tags":[14],"featured_image_src":"https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/37afede0-926b-4fb4-8352-a72d63018837.jpg","author_info":{"display_name":"importmanagerhub@sprylab.com","author_link":"https:\/\/c01.purpledshub.com\/bbcskyatnight\/author\/importmanagerhubsprylab-com\/"},"acf":{"readingTimeMinutes":"6","apple_news_title":""},"uagb_featured_image_src":{"full":["https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/37afede0-926b-4fb4-8352-a72d63018837.jpg",673,650,false],"thumbnail":["https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/37afede0-926b-4fb4-8352-a72d63018837-150x150.jpg",150,150,true],"medium":["https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/37afede0-926b-4fb4-8352-a72d63018837-300x290.jpg",300,290,true],"medium_large":["https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/37afede0-926b-4fb4-8352-a72d63018837.jpg",673,650,false],"large":["https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/37afede0-926b-4fb4-8352-a72d63018837.jpg",673,650,false],"1536x1536":["https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/37afede0-926b-4fb4-8352-a72d63018837.jpg",673,650,false],"2048x2048":["https:\/\/c01.purpledshub.com\/uploads\/sites\/77\/2022\/02\/37afede0-926b-4fb4-8352-a72d63018837.jpg",673,650,false]},"uagb_author_info":{"display_name":"importmanagerhub@sprylab.com","author_link":"https:\/\/c01.purpledshub.com\/bbcskyatnight\/author\/importmanagerhubsprylab-com\/"},"uagb_comment_info":0,"uagb_excerpt":"How a simple formula gives you a deeper understanding of the physical form of lunar 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