opendatalab/WebMainBench huggingface dataset:
# somebody help me ...

Hi everyone,

I'm having trouble with loading Max packages. I have two "Max 8" folders on my system, and I've tried placing the downloaded Max package in both locations, but I'm still unable to load subpatches. Does anyone know what might be going wrong or have any suggestions on how to resolve this issue?

Thanks in advance!

  * did you not forget to show paths to this 2 folders ?

On Mac packages get installed into ~/Documents/Max 8/Packages folder

when added using package manager.

If that downloaded package does not get recognised,

maybe it is not formatted as such 

  * it was installed on downloads folder in Mac... is packages be into /Documents/Max 8/Packages folder automatically when it is downloaded?

  * no, only if you use max package manager.

move unzipped **packages** into ~/Documents/Max 8 .... Packages

few externals in a folder don't make valid package which gets recognised as such

  * I did it!! thanks :D
### Latest Fishing Report

**Stream:** | **Letort Spring Run**  
  
  
---|---  
  
**Date of Update:** | 8/5/2020  
  
  
---|---  
  
**Author:** | TCO Boiling Springs Staff  
  
  
---|---  
  
**Highlight:** | Ants, beetles during the day and small caddis in the evening. Don't forget that cressbugs inhabit the stream year-round and are always a good bet subsurface. Slowly strip a sculpin pattern between the cress beds for bigger fish.  
  
  
---|---  
  
**Stream Level:** | Steady  
  
  
---|---  
  
**Water Temp:** | 55 degrees  
  
  
---|---  
  
**Clarity:** | Clear  
  
  
---|---  
  
**Fishing Overview:** | Summer is the time for terrestrials. Look for them after the sun has been up for a while and cast along the bank under overhanging brush.  
  
  
---|---  
  
**Specifics and Tips:** | BWO's on cloudy days. Small nymphs are producing results in the film. Try small caddis at dusk.  
  
  
---|---  
  
**Hatches:** | Sow Bugs sz. 12-20  
Midges sz. 20-26  
DeviAnt sz. 10-14  
  
  
---|---  
  
**Hot Patterns:** | TCO Cress Bugs - sz. 14-18  
TCO Scuds- sz. 14-16  
  
  
---|---  
  
**Other Information:** | TCO Boiling Springs hours: Monday - Friday: 9am - 7pm  
Saturday: 9am - 5pm  
Sunday : 9am - 4pm   
  
Click here to check out our Fly Fishing Educational Programs  
  
  
---|---  
  
**Sale Items:** | **TCO Fly Shop Sale Page** : Shirts / Jackets / Packs and more... 30%-50% off!! Check it out now  
  
  
---|---  
  
**Conservation Note:** | The big storm missed us, but we did receive a slight amount of rain. Look for some better fishing today, as things may have “freshened” Up a bit.  
  
  
---|---  
  
**General:**  
---
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#  December 2021 

Sunday | Monday | Tuesday | Wednesday | Thursday | Friday | Saturday  
---|---|---|---|---|---|---  
28 |  29 |  30 |  1 |  2 |  3 |  4  
5 |  6 |  7 |  8 |  9 |  10

  * Screening; School in Session
    * Events

|  11  
12 |  13 |  14

  * Site Council 5 PM  –  6 PM 
    * Events

|  15 |  16 |  17

  * End of Second Quarter
    * Events
  * Holiday Parties
    * Events

|  18  
19 |  20

  * No School; Teacher Work Day
    * Events

|  21

  * No School; Winter Break
    * Events

|  22

  * No School; Winter Break
    * Events

|  23

  * No School; Winter Break
    * Events

|  24

  * No School; Winter Break
    * Events

|  25  
26 |  27

  * No School; Winter Break
    * Events

|  28

  * No School; Winter Break
    * Events

|  29

  * No School; Winter Break
    * Events

|  30

  * No School; Winter Break
    * Events

|  31

  * No School; Winter Break
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|  1
##  串焊機皮帶 

  
  


本公司串焊機皮帶包括雙層復合4排孔串焊機皮帶、6排孔串焊機鋸齒接頭復合皮帶、粗糙透氣型齒接打孔串焊機皮帶、太陽能電池片串焊機皮帶等多種規格型號皮帶。 

無接縫串焊機皮帶采用軍工用的凱芙拉（芳綸）或進口高強度玻璃纖維紗經圓織設備編織筒狀織物，涂覆特氟龍乳液制成。輸送帶接口處不易斷裂，連接處周長走行穩定性好、無跑偏現象。無接口的皮帶比其他類型的串焊機皮帶使用壽命長達3倍以上。我公司生產的串焊機皮帶具有直線度與平整度高，孔位置精度高，延伸率低，熱變形小，耐高溫抗疲勞性能。 

#####  串焊機皮帶主要特點 

雙層復合串焊機皮帶采用高強度玻璃纖維紗或凱芙拉（芳綸）編織成玻璃纖維布或凱芙拉布，采用獨特的工藝技術浸漬特氟龍樹脂（PTFE）制成的。接頭處平整，走行穩定性好、不易跑偏。雙層復合串焊機皮帶接口采用獨特的齒狀連接，更有效的增強了輸送帶的強度，目前我們所做的雙層復合的串焊機皮帶里層為黑色的特氟龍耐高溫布，外層為棕色的特氟龍高溫布進過高溫熱壓成型后在通過國外進口的打孔儀器進行精密的打孔，我公司生產的串焊機皮帶還應用于國外太陽能電池片串焊機的設備上，如梅耶博格的somont，西班牙GOROASBEL等硅片自動串焊機 

目前國內大部分大型太陽能電池生產廠商均進口歐美先進的自動串焊機（例如西班牙串焊機），打孔傳送帶也完全依賴于進口，成本高，周期長，不能在第一時間滿足客戶需要，我公司在聯合國內幾家太陽能電池片串焊機設備廠家，經過一年多的努力，成功開發出太陽能硅片自動串焊機特氟龍串焊機皮帶。 

太陽能電池片串焊機皮帶主要特點：具有直線度與平整度高，孔位置精度高，延伸率低，熱變形小，耐高溫抗疲勞性能。我們可按照客戶提供的樣帶和圖紙進行定制。 

友情提醒：由于每個客戶的串焊機皮帶要求不一樣，因此在打孔的之前我們需要開設模具，所以第一次交貨期有點偏長，如急要的客戶請提前定制，以免耽擱貴司工廠的生產進度。 

版權所有： 江蘇晨昇高分子材料有限公司地址：江蘇省泰州市高港區永安洲標準廠房區6號標準廠房  技術支持：創意在線  


電話：13852613626 傳真：0523-86927446 24小時咨詢電話：13852613626 網址： www.zxgfz.com

公司專業生產： 特氟龍粘合機帶、特氟龍微波輸送帶、特氟龍網格輸送帶、特氟龍高溫浸漬布、特氟龍膠帶

蘇公網安備 32128302000354號
# BE
关于开展第十届“江苏大学十佳青年学生”评选活动的通知 各学院团委、各研究院（中心、所）团总支： 为在青年学生中树立先进典型，鼓励青年学生奋发成才，引导青年学生树立社会主义核心价值观，经研究，决定开展第十届“江苏大学十佳青年学生”评选活动。具体事项通知如下： 一、评选宗旨 开展“江苏大学十佳青年学生”评选活动，旨在树立先进典型，培养思想先进、学业优异、敢于创新、能力多样、素质全面的优秀大学生，激励广大学生践行青年使命，为实现中国梦贡献青春智慧。 二、评选机构 成立评选活动组织委员会，负责活动的组织和领导工作。组委会成员如下： 主任：缪子梅 委员：李战军 李洪波 戈国元 尹志国 金丽馥 许文荣 全 力 高 静 陈文娟 徐邦芪 组委会下设办公室，负责评选活动的具体工作。成员如下： 办公室主任：冯 琳 秘书：沈阳阳 三、评选对象 江苏大学全日制在校学生（不含在职研究生和毕业班学生）。 四、评选条件 江苏大学十佳青年学生的评选标准为：拥护党和国家的路线、方针、政策，认真学习邓小平理论和“三个代表”重要思想，贯彻落实科学发展观，德智体美等全面发展，在自强励志、创新创业、社会实践、志愿服务、社会工作、文体艺术等方面表现出色，并取得突出成绩。在上述基础上，对具备下列条件的候选人予以优先考虑： 1.校级及以上“优秀学生干部”、“三好学生”、“优秀共青团员”、“优秀团干部”、“优秀学生干部”等荣誉称号获得者。 2.在“挑战杯”全国、全省大学生课外学术科技作品竞赛和创业计划大赛、全国大学生节能减排大赛、全国大学生数学建模竞赛等重要竞赛活动中取得优异成绩者。 3.学术科研成果具有较高的理论价值和推广价值，或应用于实践领域产生显著社会经济效益者。 4.利用自身所学知识，敢于探索，勇于实践，在创新创业过程中具有一定的榜样示范作用。 5.积极参加社会实践、志愿服务活动，在社会主义精神文明与和谐社会建设中做出突出贡献者。 6.在省级及以上重大文体赛事中取得突出成绩者。 7.在逆境中顽强拼搏、自立自强且品学兼优者。 五、评选程序 1．申报阶段：（4月2日前） 各单位根据评选要求,开展相关的评选活动，在此基础上向组委会推荐候选人1—5名。候选人上报前须在本单位内通过适当的形式进行公示，并严格按照要求准备材料，按时上报。 2．预评阶段：（4月3日—4月9日） 组委会办公室对各单位推荐人选的材料进行审核，将符合要求的人选提交组委会；组委会对候选人初步评选,确定 60 名人选作为预备候选人。 3．网上投票阶段：（4月10日—4月15日） 在PU平台公布预备候选人及其事迹，进行网上投票。结合预评情况和投票结果确定20名人选作为正式候选人。 4．公示阶段：（4月16日— 4月18日） 在PU平台、江大青年网站等公布正式候选人及其事迹,进行为期3天的公示。 5．终审阶段：（4月25日前） 组委会在听取各方反馈意见的基础上，组织评委对候选人进行公开面试答辩，答辩要求准备5分钟陈述（可制作PPT）并回答评委提出的问题，最终由评委团进行综合评判，确定第十届“江苏大学十佳青年学生”获得者，并授予“江苏大学十佳青年学生”荣誉称号。 六、评选要求 1．各相关单位团委（团总支）要高度重视，认真组织，营造氛围，做好宣传发动和候选人推荐工作，确保活动顺利开展。 2．在推荐候选人的工作中，要严格按照评选条件，坚持择优原则，所推荐的学生应是本单位中最具代表性的学生。 3．申报材料应包括以下内容： （1）认真填写推荐登记表1份，并张贴二寸免冠照片； （2）被推荐人2000字的事迹材料，用A4纸4号仿宋字体统一打印； （3）提供事迹材料中涉及到的主要奖项的证书复印件和证明材料各1份。 请各相关单位团委（团总支）严格按照申报时间要求，将以上材料上交组委会办公室。 联系人：沈阳阳 联系电话：88780039 地址：校团委办公室（学工楼409） 第十届“江苏大学十佳青年学生” 评选活动组织委员会 2014年3月25日  
---
# New Home Update

|  | 

## 

## **What’s Going On?**  
  
---  
  
We know you’ve been reading and hearing a lot these days about the plans for our new building, so we wanted to send you a special informational email with “the real scoop.” **History** Paul and Renee Snider, local philanthropists and car lovers, approached us a few years ago about locating a new museum they wanted to open near us in the Docks Area or perhaps even co-locating in the same building, as they knew we were looking for a new facility/location. We looked in the McClellan Park area, Richards Blvd and elsewhere before finally striking a deal with the City to purchase our current property and build a new facility at our existing location. The Sniders generously committed to paying for both the property and the cost of construction of the new 180,000 square foot facility that would house both the California Automobile Museum as well as the new Natural History Museum. The current plan is for the Auto Museum to occupy approximately 2/3 of the space (100,000 square feet) and the Natural History Museum to occupy approximately 50,000 square feet. We will then have shared spaces such as bathrooms, elevators, foyers, event areas, etc. **Currently** Obviously, this is a huge gift for us, and we will only need to be responsible to fund the tenant improvements we decide to make to create an auto museum atmosphere. They will lease the building to us for 99 years at $1/year, and we are currently working on an Operating Agreement between the two museums. This is the kind of offer that comes along once in a lifetime. The Sniders are willing to donate $15 million+ toward a new home for the Auto Museum and their collection – an offer most museums only dream about! The Sniders have hired a team of people to work on the project, including an architect, developer, museum designer, public relations, urban planner and more. We had hoped to be able to pull permits by the end of the year, but after presenting our plan to the City Planning & Design Commission, are now realizing the process will take much more time than that, so we are slowing down our timetable. During this “slow-down,” we are working fast and furiously to do community outreach, make changes to the current plan based on recommendations from the Planning and Design Commission, and develop a more comprehensive strategy to get us through the minefield of challenges that come with a project such as this. **Timeline** Our plan is to work on the interior and exterior of the facility during the next few weeks, continue to meet with City staff to coordinate the project, do a lot of community outreach to various groups like Sacramento Convention and Visitors Bureau, Metro Chamber, Broadway Partnership, Downtown Partnership, Sacramento Animal Shelter, and more. If you would like to help, we have listed some opportunities to the right. We will be continuing to update you via email, and you can also visit the Project Website for further information.  
---
#  How a Construction Contractor Improved My Home 

For many years, I wasn't really happy with the property I was living in. I thought that the kitchen was too small. I hated trying to cook family meals as it was super stressful in such a tiny space. I used to dream about selling the house and buying somewhere new, but I didn't really have the money and I didn't want to move area with my kids still in school. My friend suggested that I hire a team of construction contractors and ask them to extend the kitchen space out into the backyard. The contractors were great. They demolished one wall and then laid a new foundation for the extension. Next, the walls were constructed and then the roof was installed. I am so happy with my new kitchen.
# L2 regularization intuition

A nice intuition for L2 regularization comes from having a prior on the distribution of parameters: the prior assumes that the parameters are close to zero. Let’s assume that the prior is $\mathcal{N}(0, \Sigma)$. The MAP estimate of the parameters would then be

$$\begin{align} \theta_{\text{MAP}} &= \text{argmax}_{\theta} ; P(\theta | D) \\\ &= \text{argmax}_{\theta} ; P(D | \theta) P(\theta) \\\ &= \text{argmax}_{\theta} ; \log P(D | \theta) + \log P(\theta) \end{align}$$

$\log P(D | \theta)$ is simply the log-likelihood of the model, and optimizing that would give you the MLE parameters. However, incorporating $\log P(\theta)$ gives us the L2 regularization parameter

$$\log P(\theta) = -\log((2\pi)^{n/2}|\Sigma|^{1/2}) - \frac{\theta^T \Sigma^{-1} \theta}{2}$$

The constant at the start goes out of the argmax, and we’re left with the L2 regularization term. Taking a negative on both sides would give us the negative log-likelihood, which is the unregularized loss function. We’d then minimize $\theta$. The net expression would look something like this:

$$\theta_{\text{MAP}} = \text{argmin}_{\theta} ; - P(D | \theta) + \frac{1}{2} \theta^T \Sigma^{-1} \theta$$

Regularization generally features a strength term $\lambda$: We can think of $\lambda$ as being the inverse of every term in the diagonal of the covariance matrix (if it is a diagonal covariance matrix). We’d then get

$$\theta_{\text{MAP}} = \text{argmin}_{\theta} ; - P(D | \theta) + \frac{1}{2} \lambda\theta^T\theta$$

And this is the familiar L2 reglarized loss function.

I got this from one of the Deep Learning lectures; it’s a nice treatment which I couldn’t find in other places (maybe because I haven’t looked hard enough). Goodfellow certainly doesn’t feature it though.
# 第十届新华网思客年会（山西）将于1月13日在临汾举行

由新华网、新华社山西分社、中共临汾市委、临汾市人民政府主办的第十届新华网思客年会（山西）将于1月13日举行。届时，刘世锦、李兆前、龚克、刘尚希、张燕生、吕廷杰、纪连海、张志安、高瑞东等知名专家学者将齐聚山西临汾，共话“中国经济的下一程”。
校園生活

  * ### 院團委

院團委是在學院黨委和上級團組織的領導下，圍繞學院黨團中心工作，結合團員青年特點，肩負團結青年、引領青年、服務青年的青年組織。

  * ### 學生團體

學生會是學生自治的重要機構,學生會本著“自我管理，自我教育，自我服務”的原則，不斷加強自身隊伍建設，不斷創新， 廣泛開展各類學生活動。

  * ### 校園風光

辦學20多年來，德州科技職業學院堅持正確的辦學方向，依法辦學、創新發展，深化高職教學改革，深入推進校企合作， 不斷完善教學設施建設

  * ### 文體活動

學校積極推進高雅文化進校園戰略，文藝活動常規化、系列化、持續化，每年開展迎新生晚會、元旦晚會、大學生藝術節、大合唱比賽、校園歌手大賽等各類文化藝術活動100多項

  * ### 教工活動

為了提高教學水平,豐富教職工的業余生活，促進職工之間的溝通交流，增強部門的凝聚力、向心力及歸屬感。學院組織了一系列教職工活動

  * ### 學院地圖

全景校園來院路線




德州科技職業學院德州校區
    辦公室：0534-7448668
    就業處：0534-7448318
    招生辦：0534-7448566
    郵 箱：sdkjxy@263.net
    地 址：山東省禹城市學院街566號

德州校區

德州科技職業學院青島校區
    辦公室：0532-58763636
    招生處：0532-58763663
    就業處：0532-58763658
    郵 箱：dekeqd@163.com
    地 址：山東省即墨市藍村鎮

青島校區團委
| |  Cultura en declive, pero no desaparecida   
---  
|  Jóvenes de la etnia Qiang mientras practican un juego tradicional. Cuatro municipios de la provincia de Sichuan, en el suroeste de China, han solicitado recientemente que uno de sus festivales tradicionales se incluya en la lista de patrimonio cultural intangible mundial de la UNESCO. De conseguirlo, este estatus contribuiría en gran medida a preservar lo que queda de una antigua cultura en vías de extinción. En enero de este año, la minoría qiang de las regiones de Beichuan, Maoxian, Lixian y Wenchuan solicitó a la organización internacional que ayudara a la supervivencia de su tradicional Festival Qiang Nian, que correspondería al año nuevo del citado pueblo. El festival se celebra a partir del primer día del décimo mes lunar según el calendario tradicional chino y dura normalmente de tres a cinco días, aunque en algunos poblados la celebración se extiende a diez días o más de banquetes y bailes. Durante el festival, de acuerdo con las costumbres de los qiang, se realizan actividades tales como la ofrenda de sacrificios al dios del Cielo, al dios de la Montaña y al dios del Pueblo, normalmente presididas por un chamán. El habitante más anciano del poblado es el encargado de abrir las vasijas de vino de fabricación local. Las canciones que se interpretan son sobretodo duetos de carácter romántico, y los bailes se llevan a cabo colectivamente, se acompañan de varios instrumentos locales, como las flautas, tambores y gongs. Normalmente, decenas de personas participan de los bailes, creando una atmósfera muy animada. Además de bailar, la gente joven también participa en diversas actividades deportivas, como una competición consistente en empujar un palo de madera, que supone para el ganador un regalo en forma de vino ofrecido por las muchachas que están contemplando la competición.  
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A long time ago, I gave my friend (Let’s call him C) an integral to solve. He came back to me a few days later and the conversation went something like this:

> **C:** Debu, I couldn’t solve that African integral you gave me.  
> **Me:** Ok, but why are you calling it African?  
> **C:** Because Africa is a really big continent, y’know?  
> **Me:** Ok, but isn’t Asia bigger?  
> **C:** Debu, please stop making me feel dumb  
> 

The integral in question appears prominently on the YG file cover as well, and is also featured in Advanced Problems in Mathematics for JEE(A) by Vikas Gupta (Question 24 chapter 7, for those following along)

> Evaluate $$\int e^{x\sin x + \cos x} \left( \frac{x^4\cos^3x - x\sin x + \cos x}{x^2 \cos^2x}\right) dx$$

This looks fearsome, but we can solve it easily by extending the trick $\int e^x (f(x) + f’(x)) dx = e^xf(x) + c$ just a little bit. Consider the function $$e^{g(x)}f(x)$$ Differentiating said function, we see that $$\frac{d}{dx} (e^{g(x)}f(x)) = e^{g(x)}(f’(x) + g’(x)f(x))$$ Rearranging the terms and integrating on both sides: $$e^{g(x)}f(x) = \int e^{g(x)}(f’(x) + g’(x)f(x)) dx$$ For the given problem, $g(x) = x\sin x + \cos x \implies g’(x) = x\cos x$. Therefore, the answer will be something of the form $$I = e^{x\sin x + \cos x} f(x) + c$$ The problem now becomes finding $f(x)$, which we can do through some rearrangement. Rewriting the problem to fit in $g’(x)$, the problem becomes: $$I = \int e^{x\sin x + \cos x} \left( x\cos x \cdot x + \frac{\sec x}{x^2} - \frac{\tan x \sec x}{x}\right) dx$$ We can see that the first term in the bracket implies that $f(x) = x$, but $f’(x) \ne \frac{\sec x}{x^2} - \frac{\tan x \sec x}{x}$. It’s easy to see that the last two terms have some kind of pattern going on, as $\frac{d}{dx}\sec x = \sec x \tan x$ and $\frac{d}{dx} \frac{1}{x} = -\frac{1}{x^2}$. The last two terms can be rewritten as $\frac{d}{dx} \left( \frac{\sec x}{x}\right)$. Go through the above steps once again if you’re unsure of what is happening, because this is important. The integral can then be rewritten as: $$I = \int e^{x\sin x + \cos x} \left( x\cos x \cdot x - \frac{d}{dx} \left( \frac{\sec x}{x}\right) \right) dx$$

We’re finally getting somewhere! We had initially assumed that $f(x) = x$, but now, seeing the differential term in the bracket, let’s assume $$f(x) = x - \frac{\sec x}{x} \implies f’(x) = 1 - \frac{d}{dx} \left( \frac{\sec x}{x}\right)$$

Also, $g’(x)f(x)$ then becomes $$g’(x)f(x) = x\cos x \left(x - \frac{\sec x}{x}\right) = x\cos x \cdot x - 1$$

Adding them together, we get $$g’(x)f(x) + f’(x) = x\cos x \cdot x - \frac{d}{dx} \left( \frac{\sec x}{x}\right)$$

Which is the term in the brackets in the integral. Thus, $f(x) = x - \frac{\sec x}{x}$. We have now solved our integral, and the answer is $$\boxed{I = e^{x\sin x + \cos x} \left( x - \frac{\sec x}{x}\right) + c}$$
##  Happy easter :)

It's Easter weekend and I have been so engrossed in Fate's Revenge (now 3/4 complete) that I didn't realize the month was halfway over with! So much for the updates :) Unfortunately, my computer decided to be temperamental and I bought a new one. Now I can write, blog, tweet, review... (really, the list is endless) without having to wait ten minutes to get on each site! It's happy time, though I haven't been able to install GIMP, but thankfully one of the old systems works enough for that. I have to go back an forth between computers when I want to access the internet or GIMP - fun, fun. Hopefully this weekend will be productive.   
Happy Easter :)   


##  it's the weekend

Well, here we are. Finally, the first weekend off studying (for a while). Unfortunately, my exam is postponed until July 2, 2014; however, I've done the studying already, so a weekend of rereading prior to test day should suffice :0) I'd have rather had it over and done with but those are the breaks, right? So what's in store this weekend? Well . . . tonight is RELAXING! Tomorrow, perhaps some cleaning (so needed), and maybe catch up on all my reading that's gone on the back-burner. Then, in the evening and Sunday? WRITING! If you don't say it, I will . . . FINALLY! (Too much?) I'm happy to get back to Twisted Fate with Book 4: Fate's Revenge.   
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# lighten flywheel

Discussion in HPI Baja Engine and Exhaust forum, started by juniorbooger.   
  
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is any one running the lighten fly wheel and how do you like it thanks  
  
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I do not run it,but I beleive that it is less efficient in cooling because it has smaller fins.   
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I have lightened flywheel on two of my baja, A Darksoul CNC on my on0road and a DDM lightened one on my basher and goped. I see a big difference all the way around better low end and great top RPM. Have never had a problem with cooling even on my goped!!!  
Just one word of advice dont use the Darksoul CNC for bashing or off-road use i learned the hard way even with outerwear all the way around.  
  
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|  is any one running the lighten fly wheel and how do you like it thanks  
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They are all that I run. Never had a problem but I do not run in the sand and I typically shut down the baja after a tank to relax for alittle between tanks. I cant keep my focus for two hours at a time. 30 minute runs is good for me   
  
Mine is lightened. Mine feels pretty good. But I am in a colder climate and mine sees alot of run time in the snow. So I know nothing about cooling issues.   
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Another note, there are still cooling fins on the opposite side of the flywheel, so there is still a lot of cooling air circulating.   
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i run a lightened flywheel on one of my cars. i lightened it my self, i looked at a picture of dark souls wheel and did it just the same way. it does make a difference on spool up an top rpms, the throttle response is alot better. i use a pullstart cover and crankcase shroud cover. i think the crankcase outerwear is key to not exploding one of these. i run in rocks and whatever never had a problem.   
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|  I have lightened flywheel on two of my baja, A Darksoul CNC on my on0road and a DDM lightened one on my basher and goped. I see a big difference all the way around better low end and great top RPM. Have never had a problem with cooling even on my goped!!!  
Just one word of advice dont use the Darksoul CNC for bashing or off-road use i learned the hard way even with outerwear all the way around.  
  
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Even with the mesh mod?
# Limit involving higher order infinitesimals

Simple limit problems consist of the form $\lim_{x \to 0}\frac{O_1(x)O_2(x)..}{O_a(x)O_b(x)..}$, such as $\lim_{x \to 0} \frac{\sin 3x \tan 2x \tan^{-1} 5x}{x^2 \ln(1+x)}$. Here, the infinitesimals are well defined and cancel out easily. Some trickier limit problems involve the difference of two infinitesimals. A good example is $\lim_{t \to 0} \frac{\sin t - \tan t}{t^3}$. These kind of problems require expansions to solve them cleanly. L’Hopital rule is tedious as it will need to be repeated atleast thrice for this given limit, which involves infinitesimals of order 3. To illustrate how to solve these problems, I’ll give a trickier example:

> Evaluate the limit $$\lim_{n \to \infty} \left( e - \left( 1 + \frac1n \right) ^n \right) ^\frac 1n$$

It’s easy to assume that the limit $\lim_{n \to \infty} \left( 1 + \frac1n \right)^n$ evaluates to $e$, which means that the expression inside the outer brackets would tend to zero, giving the answer as zero. However, all is not as it seems. This is a good example of difference between infinitesimals yielding a higher order infinitesimal. To start with, consider the function $$f(n) = \lim_{n \to \infty} \left( 1 + \frac1n \right)^n$$ Taking a logarithm both sides and using a taylor series expansion for the log on the right side gives $$\begin{gather} \log f(n) = \lim_{n \to \infty} n \left( \frac1n - \frac 1{2n^2} + O\left(\frac 1 {n^3}\right) \right) \\\ f(n) = \lim_{n \to \infty}e^{1 - \frac 1{2n} + O\left(\frac 1 {n^2}\right)} \end{gather}$$

If we substitute $f(n)$ in the limit now, we can see that $$L = \lim_{n \to \infty} \left( e \left( 1 - e^{-\frac 1{2n} + O\left(\frac 1 {n^2}\right)}\right) \right) ^\frac 1n$$

Taking a logarithm both sides gives us $$\log L = \lim_{n \to \infty} \underbrace{\frac{1}{n}}_{f_1} + \underbrace{\frac{ \log\left( 1 - e^{-\frac 1{2n} + O\left(\frac 1 {n^2}\right)}\right)}{n}}_{f_2}$$

$f_1$ is trivially $0$. $f_2$ can be evaluated using L’Hopital rule, which gives $$f_2 = \lim_{n \to \infty} \frac{\frac{1}{2n^2} + … }{1 - e^{-\frac 1{2n} + O\left(\frac 1 {n^2}\right)}}$$

This evaluates to $0$ as the numerator is a higher order infinitesimal than the denominator. Thus, we have $$\log L = 0$$

Taking an exponent both sides gives us $$\boxed{L = 1}$$